NEWTON’S FINAL REVIEW

EXAMPLE 1:

A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.

Solution:

  Figure 4.1: Problem 4.1

a) From the 2nd Law:

ma = T - mg   a =

- g

  

  =

- 9.8 m/s 2 = 3.8 m/s 2

(5)

b) Since a is constant

v = v0 + at     = 0 + 3.8(2) = 7.6  m/s. (6)

EXAMPLE 2:

A hockey puck of mass .5 kg travelling at 10 m/s slows to 2.0 m/s over a distance of 80 m. Find a) the frictional force acting on the puck and b) the coefficient of kinetic friction between the puck and the surface.

Solution:

  Figure 4.2: Problem 4.2

a) First we find the acceleration of the puck from the kinematic equations of motion. We have, v0 = 10  m/s , v = 2  m/s and x = 80  m . The third equation of motion gives,

v 2 = v02 + 2ax   

a

=

= = - 0.6  m/s 2

(7)

From the Second Law: In the x-direction,

fk = ma = .5(- 0.6) = - 0.3  N.      (8)b)

Use fk = - N . From the y-component of the 2nd Law: N - mg = 0 . Combining,

- mg= ma   

= - a/g   

  = 0.061. (9)

EXAMPLE 3:

A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an

elevator. Assume that the scale reads in newtons. Find the scale reading when the elevator is a) accelerating upward at .5 m/s 2, b) going up at a constant speed of 3.0 m/s and c) going up but decelerating at 1.0 m/s 2.

Solution:

  Figure 4.3: Problem 4.3

From the 2nd Law:

Fs - mg = ma   

Fs

= m(g + a). (10)

This gives: a)

Fs = 50(9.8 + 0.5) = 515  N b)

Fs = 50(9.8 + 0) = 490  N c)

Fs = 50(9.8 - 1.0) = 440  N

EXAMPLE 4:

A wooden plank is raised at one end to an angle of 30 o . A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The

coefficient of kinetic friction between the box and the plank is = 0.20 . Find a) the rate of acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero.

Solution:

  Figure 4.4: Problem 4.4

a) Find the components of the weight of the object:

Fxg = - mgsin   Fyg = - mgcos .   

Write out the two components of Newton's 2nd Law:

  x : - mgsin + fk = - ma     y : N - mgcos = 0. (11)

Using fk = N we get,

ma = - (mgcos ) + mgsin

  

a

= g(sin - cos )

  

  = 9.8(sin 30 - 0.2cos 30) = 3.20 m/s 2 (12)b)

Since a is constant and v 2 = v02 + 2ax . With x = 1  m , v0 = 0 we have

v = = 2.53  m/s .

       

EXAMPLE 5:

A constant horizontal force, Fapp of magnitude 25.0 N is applied to block A of mass mA = 4.5 kg, which pushes against block B of mass mB = 6.5 kg. The blocks slide over a frictionless surface, along an x axis. (a) What is the acceleration of the blocks?(b) What is the force FBA on block B from A?

i. Draw a labeled free-body diagram for the two boxes during the time when the boxes are accelerating, specifying all the forces acting on the block. (Be sure to specify the type of force and the object causing each force.) Wherever you can, compare the magnitudes of forces.

Recall action-reaction forces : box A must be pushing on box B with the same force as box B is pushing on box A.

ii. Write equations for the free body diagrams for box A and box B:

Box B: FnetB = FBA

Box A: FnetA = Fapp – FAB

iii.

NOTE: The force net of each box is the total sum force on the box which is its mass times its acceleration.

FnetB = mAaA

FnetA = mBaB

The boxes must have identical acceleration otherwise the boxes would not be connected. aB = aA

Unfortunately, we do not know each Fnet, therefore we must substitute into the other equation to find the acceleration of box A and B. Note: the acceleration of the boxes is equal.

FnetA = Fapp – FAB

substituting for FAB (FAB is = but directionally opp to FBA)FnetA = Fapp – FnetB

mA a = Fapp - mB a Fapp = (mA+ mB)aa = Fapp/(mA+mB) = 25.0/(4.5 + 6.5) = 2.2727 = 2.27 m/s2

b. What is the force FBA on block B from block A?

i. FBA = mBa = 6.5*2.2727 = 14.8 N The acceleration on block A and B are the same.

EXAMPLE 6:

Determine the time it will take the mass to fall a distance of 1.5 meters if starting from rest with a coefficient of kinetic fiction of 0.015.

 

NEWTON’S REVIEW

1. Which of the following statements are true of inertia? List all that apply.

a. Inertia is a force. b. Inertia is a force which keeps stationary objects at rest and moving objects in

motion at constant velocity. c. Inertia is a force which brings all objects to a rest position. d. All objects have inertia. e. A more massive object has more inertia than a less massive object.

f. Fast-moving objects have more inertia than slow-moving objects. g. An object would not have any inertia in a gravity-free environment (if there is

such a place). h. Inertia is the tendency of all objects to resist motion and ultimately stop. i. In a gravity-free environment (should there be one), a person with a lot of inertia

would have the same ability to make a turn as a person with a small amount of inertia.

2. Which of the following statements are true of the quantity mass? List all that apply.

a. The mass of an object is dependent upon the value of the acceleration of gravity. b. The standard metric unit of mass is the kilogram. c. Mass depends on how much stuff is present in an object. d. The mass of an object is variable and dependent upon its location. e. An object would have more mass on Mount Everest than the same object in the

middle of Lake Michigan. f. People in Weight Watcher's are really concerned about their mass (they're mass

watchers). g. The mass of an object can be measured in pounds. h. If all other variables are equal, then an object with a greater mass would have a

more difficult time accelerating. i. If all other variables are equal, then it would require less exerted force to stop a

less massive object than to stop a more massive object. j. The mass of an object is mathematically related to the weight of the object.

3. Which of the following statements are true of the quantity weight? List all that apply.

a. The weight of an object is dependent upon the value of the acceleration of gravity. b. Weight refers to a force experienced by an object. c. The weight of an object would be less on the Moon than on the Earth. d. A person could reduce their weight significantly by taking an airplane ride to the

top of Mount Everest. e. Two objects of the same mass can weigh differently. f. To gain weight, one must put on more mass. g. The weight of an object can be measured in kilograms. h. The weight of an object is equal to the force of gravity acting upon the object. i. When a chemistry student places a beaker on a balance and determines it to be

84.3 grams, they have weighed the beaker.

4. Which of the following statements are true of an object that experiences balanced forces (or unbalanced forces)? List all that apply.

a. If a person is moving to the right, then the forces acting upon it are NOT balanced.

b. A balance of forces is demonstrated by an object which is slowing to a stop. c. It would take an unbalanced force to keep an object in motion.

d. If an object is moving with a constant speed in a circle, then the forces acting upon the object are balanced.

e. If an object is accelerating at a constant rate of acceleration, then the forces acting upon the object are balanced.

f. It is NOT possible for just three forces to be acting upon an object and they still balance each other.

g. A free-falling object experiences a balance of forces. h. Balanced forces cause stationary objects to remain at rest and moving objects to

come to rest. i. Unbalanced forces cause objects to move.

5. Consider Newton's first law of motion to determine which of the following statements are true? List all that apply.

a. Newton's first law of motion is applicable to both moving and non-moving objects.

b. If a football is moving upwards and rightwards towards the peak of its trajectory, then there are both rightwards and upwards forces acting upon it.

c. It would take an unbalanced force to keep an object in motion. d. If an object is at rest, then there are no forces acting upon the object. e. It would take an unbalanced force to keep an object in motion at a constant

velocity. f. It is the natural tendency of all objects to eventually come to a rest position. g. A pendulum bob is set into its usual back-and-forth periodic motion. After some

time (perhaps 10 minutes), the pendulum bob comes to a rest position. This is best explained by the idea of inertia - all objects eventually resist motion.

h. If a 3-kg rock is thrown at a speed of 2 m/s in a gravity-free environment (presuming one could be found), then an unbalanced force of 6 N would be required to keep the rock moving at a constant speed.

i. It would take an unbalanced force to cause an object to accelerate from rest.

6. Consider Newton's second law of motion to determine which of the following statements are true? List all that apply.

a. If an object is accelerating to the right, the net force on the object must be directed towards the right.

b. If an object is moving to the right and slowing down, then the net force on the object is directed towards the left.

c. Accelerating objects are either slowing down or speeding up. d. The acceleration of an object is directly dependent upon its mass and inversely

dependent upon its net force. e. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is

increased by a factor of 2, then the new acceleration would be 10 m/s/s. f. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is

increased by a factor of 3, then the new acceleration would be 11 m/s/s.

g. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a factor of 2, then the new acceleration would be 16 m/s/s.

h. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a factor of 4, then the new acceleration would be 2 m/s/s.

i. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2 and the mass of the object is decreased by a factor of 2, then the two factors would offset each other and the acceleration would still be 8 m/s/s.

j. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 4 m/s/s.

k. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is decreased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 1 m/s/s.

l. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 4 and the mass of the object is increased by a factor of 2, then the new acceleration would be 16 m/s/s.

m. A 2-kg object accelerates from rest to a final velocity of 6 m/s in 3 seconds. The net force acting upon the object is 12 N.

n. A 10-kg object slows down from 24 m/s to a final velocity of 9 m/s in 3 seconds. The net force acting upon the object is 80 N.

7. Big Bubba has a mass of 100 kg on the earth. What is Big Bubba's mass on the moon where the force of gravity is approximately 2/5-th that of Earth's? ________ Explain or show your work.

8. Little Billie weighs 360 N on Earth. What is Little Billie's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ________ Explain or show your work.

9. TRUE or FALSE:

For an object resting upon a non-accelerating surface, the normal force is equal to the weight of the object.

10. Which one(s) of the following force diagrams depict an object moving to the right with a constant speed? List all that apply.

 

11. As you sit in your chair and study your physics (presuming that you do), the force of gravity acts downward upon your body. The reaction force to the force of the Earth pulling you downward is ___.

a. the force of the chair pushing you upward

b. the force of the floor pushing your chair upward

c. the force of the Earth pushing you upward

d. the force of air molecules pushing you upwards

e. the force of your body pulling the Earth upwards

f. ... nonsense! Gravity is a field force and there is no such reaction force.

12. A golf pro places a ball at rest on the tee, lines up his shot, draws back his club, and lets one rip. During the contact of the golf club with the golf ball, the force of the club on the ball is ____ the force of the ball on the club and the acceleration of the club is ____ than the acceleration of the ball.

a. greater than, greater than b. greater than, equal to c. greater than, less than

d. less than, less than e. less than, equal to f. less than, greater than

g. equal to, equal to h. equal to, greater than i. equal to, less than

Each one of Newton's Laws can play a role in any one particular situation. However, one of the laws is often most obviously dominant in governing the motion of a situation. Pick which of Newton's most governs the situations described below.

a. First Law (inertia) b. Second Law (F = m•a) c. Third Law (action-reaction)

13. A helicopter must have two sets of blades in order to fly with stability.

14. If you were in an elevator and the cable broke, jumping up just before the elevator hit the ground would not save you. Sorry.

15. You usually jerk a paper towel from a roll in order to tear it instead of pulling it smoothly.

16. A student desk changes the amount of force it puts on other objects throughout a school day.

17. Heavy objects are not easier to move around in a horizontal fashion on the Moon than on the Earth.

18. The stronger, heavier team in a tug-of-war does not create a larger tension in the rope than the weaker, lighter team.

19. When a horse pulls a wagon, the force that causes the horse to move forward is….a. the force he exerts on the ground.b. the force he exerts on the wagon.c. the force the ground exerts on himd. the force the wagon exerts on him.

Answer: c) this is the only force acting on him in the forward direction

For the next several questions, consider the velocity-time plot below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter. 

20. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.

21. During which time interval(s), if any, are the forces acting upon the object balanced.? List all that apply.

22. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.

23. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

24. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.

For the next several questions, consider the dot diagram below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter.

25. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.

26. During which time interval(s), if any, are the forces acting upon the object balanced.? List all that apply.

27. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.

28. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

29. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.

For the next several questions, consider the trajectory diagram shown below for a projectile thrown at an angle to the horizontal. The vector arrows represent the horizontal and vertical components of the projectile's velocity. Several points in the trajectory are labeled with a letter. Use the trajectory diagram to answer the next several questions. (Consider air resistance to be negligible.)

30. At which point(s), if any, are there no forces acting upon the object? List all that apply.

31. At which point(s), if any, are the forces acting upon the object balanced.? List all that apply.

32. At which point(s), if any, is there a net force acting upon the object? List all that apply.

33. At which point(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

34. At which point(s), if any, is the net force acting upon the object directed upward? List all that apply.

35. Construct free-body diagrams for the following physical situations at the instant in time for which they are described. As is always done in free-body diagrams, label the forces according to type and draw the arrows such that their length reflectshe magnitude of the force.

a. A book is being pushed to the right across a table surface with a constant velocity. (Neglect Fair.)

b. A rightward-moving box (which was previously set into rightward motion across the floor) gradually slows to a stop.

 

c. A falling skydiver has reached a terminal velocity.

 

d. A student is pushing lightly upon a large box in an attempt to push it to the right across the floor, but the box fails to move.

 

e. An elevator (held by a cable) is moving upwards at a constant velocity. (Neglect Fair.)

f. A picture hangs symmetrically by two wires oriented at angles to the vertical.

For the following problems, draw free-body diagrams and solve for the requested unknown. Use g = 9.8 m/s/s.

36. A 72-kg skydiver is falling from 10 000 feet. After reaching terminal velocity, the skydiver opens his parachute. Shortly thereafter, there is an an instant in time in which the skydiver encounters an air resistance force of 1180 Newtons. Determine the acceleration of the skydiver at this instant. PSYW

37. A 5.2-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine the acceleration of the object. (Neglect air resistance.) PSYW

38. A 921-kg sports car is moving rightward with a speed of 29 m/s. The driver suddenly slams on the brakes and the car skids to a stop over the course of 3.2 seconds with the wheels locked. Determine the average resistive force acting upon the car. PSYW

39. A tow truck exerts a 18300-N force upon a 1200-kg car to drag it out of a mud puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane of

motion of the car is horizontal. Determine the time required to drag the car a distance of 6.9 meters from its rest position. PSYW

40. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the speed (in m/s) of the bucket after 1.59 seconds. PSYW

41. A 22.6-N horizontal force is applied to a 0.071-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds.

42. An object rests upon an inclined plane. If the angle of incline is increased, then the normal force would _______.

a. increase b. decrease c. remain the same

43. Three pictures of equal weight (20 N) are hung by wires in three different orientations. In which orientation are the wires least likely to break? _______ Explain why.

44. A 50-N force is applied at an angle of 30 degrees north of east. This would be the same as applying two forces at

a. 43 N, east and 7 N, north b. 35 N, east and 15 N, north c. 25 N, east and 25 N, north d. 43 N, east and 25 N, north

Use the approximation that g=~10 m/s2 to fill in the blanks in the following diagrams. DON’T WORRY ABOUT SIG FIGS FOR QUES 45 - 51

45. 46.

 

 

 

47.

 

48.

 

49.

 

50.

 51.

 

 

52. A 945-kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.972, determine the distance required to stop. PSYW

53. A student pulls a 2.00-kg backpack across the ice (assume frictionless) by pulling at a 30.0° angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force. PSYW

 

54. Splash Mountain at Disney World in Orlando, Florida is the steepest water plume ride in the country. Occupants of the boat fall from a height of 100.0 feet (3.2 ft = 1 m) down a wet ramp which makes a 45° angle with the horizontal. Consider the mass of the boat and its occupants to be 1500 kg. The coefficient of friction between the boat and the ramp is 0.10. Determine the frictional force, the acceleration, the distance traveled along the incline, and the final velocity of the boat at the bottom of the incline. PSYW

55. At last year's Homecoming Pep Rally, Trudy U. Skool (attempting to generate a little excitement) slid down a 42° incline from the GBS dome to the courtyard below. The coefficient of friction between Trudy's jeans and the incline was 0.650. Determine Trudy's acceleration along the incline. Begin with a free-body diagram. PSYW

56. Consider the two-body system at the right. There is a 0.250-kg object accelerating across a rough surface. The sliding object is attached by a string to a 0.10 kg object which is suspended over a pulley. The coefficient of kinetic friction is 0.183. Calculate the acceleration of the block and the tension in the string. PSYW

 

57.

58. Two boxes resting on a frictionless surface are connected by a massless cord. The boxes have masses of 14.0 and 11.0 kg. A horizontal force of Fapp = 50.0 N is applied by pulling on the lighter box.What is the acceleration of each box and the tension in the cord?

REVIEW KEY1. DE2. BCFHIJ3. ABCH and possibly EF4. None of these are true, though one might make a strong argument for I.5. AI6. ABHJKL7. 100 kg8. ~36 kg9. False10. AC11. E12. I13. C14. A15. A16. C17. A18. C19. C20. NONE21. BDFH22. ACEG23. AE24. CG25. NONE26. ACEGI27. BDFH28. BF29. DH

30. NONE31. NONE32. ABCDEFG33. NONE34. NONE35. –36. 6.6 m/s2, up37. 5.0 m/ s2, right38. 8.3 x 103 N39. 6.4 s40. 14.2 m/s41. 23 m/s42. B43. 60-degree angle to the horizontal44. D45.

46.

47.

48. .

49. .

50. .

51. .

52. 26.8 m53. 0.289 N54. Ffrict = 2.1 x 103 N; a = 5.5 m/s/s; d = 44 m; vf = 22 m/s55. 1.8 m/s2

56. a = 1.5 m/s2 ; Ftens = 0.83 N57. 11 m58. 2.00 m/s2 , 22.0 N

NEWTON’S REVIEW SOLUTIONS

1. Which of the following statements are true of inertia? List all that apply.

j. Inertia is a force. k. Inertia is a force which keeps stationary objects at rest and moving objects in

motion at constant velocity. l. Inertia is a force which brings all objects to a rest position. m. All objects have inertia. n. A more massive object has more inertia than a less massive object. o. Fast-moving objects have more inertia than slow-moving objects. p. An object would not have any inertia in a gravity-free environment (if there is

such a place). q. Inertia is the tendency of all objects to resist motion and ultimately stop. r. In a gravity-free environment (should there be one), a person with a lot of inertia

would have the same ability to make a turn as a person with a small amount of inertia.

Answer: DE

a. False - Inertia is not a force.

b. False - Inertia is NOT a force.

c. False - Inertia is NOT a force. Inertia is simply the tendency of an objects to resist a change in whatever state of motion that it currently has. Put another way, inertia is the tendency of an object to "keep on doing what it is doing." Mass is a measure of an object's inertia. The more mass which an object has, the more that it sluggish towards change.

d. True - Bet money on this one. Any object with mass has inertia. (Any object without mass is not an object, but something else like a wave.)

e. True - Mass is a measure of an object's inertia. Objects with greater mass have a greater inertia; objects with less mass have less inertia.

f. False - The speed of an object has no impact upon the amount of inertia that it has. Inertia has to do with mass alone.

g. False - Inertia (or mass) has nothing to do with gravity or lack of gravity. In a location where g is close to 0 m/s/s, an object loses its weight. Yet it still maintains the same amount of inertia as usual. It still has the same tendency to resist changes in its state of motion.

h. False - Inertia is NOT the tendency to resist motion, but rather to resist changes in the state of motion. For instance, its the tendency of a moving object to keep moving at a

constant velocity (or a stationary object to resist changes from its state of rest).

i. False - Once more (refer to g), inertia is unaffected by alterations in the gravitational environment. An alteration in the g value effects the weight of an object but not the mass or inertia of the object.

2. Which of the following statements are true of the quantity mass? List all that apply.

k. The mass of an object is dependent upon the value of the acceleration of gravity. l. The standard metric unit of mass is the kilogram. m. Mass depends on how much stuff is present in an object. n. The mass of an object is variable and dependent upon its location. o. An object would have more mass on Mount Everest than the same object in the

middle of Lake Michigan. p. People in Weight Watcher's are really concerned about their mass (they're mass

watchers). q. The mass of an object can be measured in pounds. r. If all other variables are equal, then an object with a greater mass would have a

more difficult time accelerating. s. If all other variables are equal, then it would require less exerted force to stop a

less massive object than to stop a more massive object. t. The mass of an object is mathematically related to the weight of the object.

Answer: BCFHIJ

a. False - Mass is independent of the gravitational environment that an object is in and dependent solely upon the number of atoms in the object and the type of atoms (Carbon: ~12 g/mol; Hydrogen: ~1 g/mol ; Oxygen: ~16 g/mol). Because of this, mass is said to be invariable (unless of course, an object loses some of its atoms) - a constant quantity which is independent of the acceleration of gravity and therefore independent of location. (Weight on the other hand depends upon the gravitational environment.)

b. True - Know this one. Kilograms is for mass and Newtons is for force.

c. True - This is kind of a simple definition of mass but it does do the job (provided stuff means atoms or material).

d. False - See explanation to #2d.

e. False - An object has the same mass on Mount Everest as it does at sea level (or near sea level); only the weight of the object would be slightly different in these two locations.

f. True - Weight Watcher's participants only use a measurement of their weight as a reflection of how many atoms of flesh that they have burned from their bodies. Their real

interest is in losing mass for reasons related to health, appearance, etc.

g. False - Pounds is a unit of force commonly used in the British system of measurement. It is not a metric unit and it is not a unit of mass. Kilogram is the standard metric unit of mass and slug is the British unit.

h. True - Weight and force of gravity are synonymous terms. You should quickly become comfortable with the terms mass, weight and force of gravity; it will save you many headaches as we continue through the course.

i. True - A less massive object has less inertia and as such would offer less resistance to changes in their velocity. For this reason, a less massive object requires less force to bring from a state of motion to a state of rest.

j. True - The weight of an object is the mass of the object multiplied by the acceleration of gravity of the object. Mass and weight are mathematically related by the equation: Weight (or Fgrav) = m•g

3. Which of the following statements are true of the quantity weight? List all that apply.

j. The weight of an object is dependent upon the value of the acceleration of gravity. k. Weight refers to a force experienced by an object. l. The weight of an object would be less on the Moon than on the Earth. m. A person could reduce their weight significantly by taking an airplane ride to the

top of Mount Everest. n. Two objects of the same mass can weigh differently. o. To gain weight, one must put on more mass. p. The weight of an object can be measured in kilograms. q. The weight of an object is equal to the force of gravity acting upon the object. r. When a chemistry student places a beaker on a balance and determines it to be

84.3 grams, they have weighed the beaker.

Answer: ABCH and possibly EF

a. True - The weight of an object is equal to the force of gravity acting upon the object. It is computed by multiplying the object's mass by the acceleration of gravity (g) at the given location of the object. If the location of the object is changed, say from the Earth to the moon, then the acceleration of gravity is changed and so is the weight. It is in this sense that the weight of an object is dependent upon the acceleration of gravity.

b. True - This statement is true in the sense that the weight of an object refers to a force - it is the force of gravity.

c. True - The weight of an object depends upon the mass of the object and the

acceleration of gravity value for the location where it is at. The acceleration of gravity on the moon is 1/6-th the value of g on Earth. As such, the weight of an object on the moon would be 6 times less than that on Earth.

d. False - A trip from sea level to the top of Mount Everest would result in only small alterations in the value of g and as such only small alterations in a person's weight. Such a trip might cause a person to lose a pound or two.

e. Mostly True - Two objects of the same mass can weigh differently if they are located in different locations. For instance, person A and person B can both have a mass of 60 kg. But if person A is on the Earth, he will weigh ~600 N, whereas person B would weight ~100 N on the moon.

f. Kinda True (Mostly False) - Weight is the product of mass and the acceleration of gravity (g). To gain weight, one must either increase their mass or increase the acceleration of gravity for the environment where they are located. So the statement is true if one disregards the word MUST which is found in the statement.

g. False - By definition, a free-falling object is an object upon which the only force is gravity. Such an object is accelerating at a rate of 9.8 m/s/s (on Earth) and as such cannot be experiencing a balance of forces.

h. True - This statement is the precise definition of weight. Weight is the force of gravity.

i. False - This student has determined the mass of the beaker, not the weight. As such, he/she has massed the beaker, not weighed it.

4. Which of the following statements are true of an object that experiences balanced forces (or unbalanced forces)? List all that apply.

j. If a person is moving to the right, then the forces acting upon it are NOT balanced.

k. A balance of forces is demonstrated by an object which is slowing to a stop. l. It would take an unbalanced force to keep an object in motion. m. If an object is moving with a constant speed in a circle, then the forces acting

upon the object are balanced. n. If an object is accelerating at a constant rate of acceleration, then the forces acting

upon the object are balanced. o. It is NOT possible for just three forces to be acting upon an object and they still

balance each other. p. A free-falling object experiences a balance of forces. q. Balanced forces cause stationary objects to remain at rest and moving objects to

come to rest. r. Unbalanced forces cause objects to move.

Answer: None of these are true, though one might make a strong argument for I.

a. FALSE - An object which is moving to the right could have unbalanced forces, but only if it is accelerating. The presence of unbalanced forces must always be associated with acceleration, not mere motion. In this case, an object moving to the right could have a balance of forces if it is moving with a constant velocity.

b. FALSE - An object would never slow to a stop unless the forces acting upon it were unbalanced. In fact, an object which slows down must have a unbalanced force directed in the direction opposite their motion.

c. FALSE - An unbalanced force is only required to accelerate an object. A balance of forces is required to keep an object moving at a constant velocity. For instance, a car moving to the right at constant velocity encounters as much rightward force as leftward force.

d. FALSE - An object which moves in a circle has a changing direction. As such, there is an acceleration and this acceleration requires that there be an unbalanced force present on the object.

e. FALSE - Any object that accelerates has a changing velocity. An object that accelerates at a constant rate has a velocity that changes by the same amount each second. For instance, a free-falling object changes its velocity by -9.8 m/s ever second. It is said to have a constant acceleration of -9.8 m/s2. A free-falling object, or any object with an acceleration (whether constant or non-constant) must be experiencing an unbalanced force.

f. FALSE - Consider an object which weighs 1000 N (a 1000 N downward force) which is being pulled on by two people, each exerting 500 N of upward force. Such an object has three forces acting upon it and the three forces together balance each other.

g. FALSE - A free-falling object is an object upon which the only force is gravity. As such, there is an unbalanced force acting upon it; this unbalanced force explains its acceleration.

h. FALSE - Balanced forces cause stationary objects to stay at rest. However balanced forces would never cause moving objects to stop; an unbalanced force would be required to stop a moving object.

i. FALSE - Unbalanced forces do more than cause objects to move; unbalanced forces cause objects to accelerate. Though one could make a strong argument that an object that is accelerating must also be moving (albeit with a changing velocity). In this sense, this statement is true.

5. Consider Newton's first law of motion to determine which of the following statements are true? List all that apply.

j. Newton's first law of motion is applicable to both moving and non-moving objects.

k. If a football is moving upwards and rightwards towards the peak of its trajectory, then there are both rightwards and upwards forces acting upon it.

l. It would take an unbalanced force to keep an object in motion. m. If an object is at rest, then there are no forces acting upon the object. n. It would take an unbalanced force to keep an object in motion at a constant

velocity. o. It is the natural tendency of all objects to eventually come to a rest position. p. A pendulum bob is set into its usual back-and-forth periodic motion. After some

time (perhaps 10 minutes), the pendulum bob comes to a rest position. This is best explained by the idea of inertia - all objects eventually resist motion.

q. If a 3-kg rock is thrown at a speed of 2 m/s in a gravity-free environment (presuming one could be found), then an unbalanced force of 6 N would be required to keep the rock moving at a constant speed.

r. It would take an unbalanced force to cause an object to accelerate from rest.

Answer: AI

a. TRUE - Absolutely true. Like all true scientific laws, they govern all objects. In the case of Newton's first law of motion: An object that is nonmoving remains at rest (unless acted upon by an unbalanced force); and a moving object will continue in its motion at a constant velocity (unless acted upon by an unbalanced force).

b. FALSE - A football which is moving upwards and rightwards towards its peak, then it has both an upward and a rightward velocity; it does not however have an upward and a rightward force. In fact, if acting as a projectile, it has no horizontal force and maintains a constant horizontal velocity; similarly, it would have a downward force of gravity and a slowing down motion as it rises. If the football were not a projectile, then the horizontal force would be leftward (air resistance opposing its motion) and the vertical force would be gravity and air resistance, both directed downward.

c. FALSE - An unbalanced force would accelerate an object. If directed against its motion, then it would actually slow it down rather than keep is motion going. A balance of forces is all that is required to keep an object going at a constant velocity. An unbalanced force directed in the direction of motion would be required to keep an object going with an increasing speed.

d. FALSE - If an object is at rest, then there are no unbalanced forces acting upon it. There is a force of gravity and at least one other upward force capable of balancing the force of gravity.

e. FALSE - This is dead wrong. It would take a balance of forces to keep an object in motion at constant velocity. An unbalanced force would cause some form of acceleration.

f. FALSE - If you answered TRUE, then Galileo and Newton just rolled over in their

grave. It is the natural tendency of all objects to maintain their velocity and to resist changes in whatever state of motion that they have. This is the law of inertia.

g. FALSE - All objects resist changes in their state of motion. In the absence of unbalanced forces, they maintain their velocity (whether zero or nonzero). The pendulum changes its state of motion due to an unbalanced force - the force of air resistance.

h. FALSE - For an object to maintain a constant velocity, 0 Newtons of net force (i.e., a balance of forces) is required.

i. TRUE - Unbalanced forces cause stationary objects to accelerate from rest. In the absence of an unbalanced force, a stationary object would remain at rest.

6. Consider Newton's second law of motion to determine which of the following statements are true? List all that apply.

o. If an object is accelerating to the right, the net force on the object must be directed towards the right.

p. If an object is moving to the right and slowing down, then the net force on the object is directed towards the left.

q. Accelerating objects are either slowing down or speeding up. r. The acceleration of an object is directly dependent upon its mass and inversely

dependent upon its net force. s. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is

increased by a factor of 2, then the new acceleration would be 10 m/s/s. t. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is

increased by a factor of 3, then the new acceleration would be 11 m/s/s. u. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a

factor of 2, then the new acceleration would be 16 m/s/s. v. An object has an acceleration of 8 m/s/s. If the mass of the object is increased by a

factor of 4, then the new acceleration would be 2 m/s/s. w. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is

increased by a factor of 2 and the mass of the object is decreased by a factor of 2, then the two factors would offset each other and the acceleration would still be 8 m/s/s.

x. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 4 m/s/s.

y. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is decreased by a factor of 2 and the mass of the object is increased by a factor of 4, then the new acceleration would be 1 m/s/s.

z. An object has an acceleration of 8 m/s/s. If the net force acting upon the object is increased by a factor of 4 and the mass of the object is increased by a factor of 2, then the new acceleration would be 16 m/s/s.

aa. A 2-kg object accelerates from rest to a final velocity of 6 m/s in 3 seconds. The net force acting upon the object is 12 N.

bb. A 10-kg object slows down from 24 m/s to a final velocity of 9 m/s in 3 seconds. The net force acting upon the object is 80 N.

Answer: ABHJKL

a. TRUE - The acceleration is directly related to the net force and the direction of the acceleration is always the same as the direction of the net force. When it comes to force, objects can be thought of as being in the middle of a tug-of-war between the individual forces. The force that wins the tug-of-war is the force which determines the direction of the acceleration. So if a rightward force wins over a leftward force, the acceleration will be to the right.

b. TRUE - An object which is slowing down has an acceleration which is directed opposite the motion of the object. So an object which moves to the right and slows down experiences a leftward acceleration and therefore a leftward net force.

c. FALSE - Acceleration involves a change in velocity and velocity is a vector with a magnitude (15 m/s, 22 m/s, etc.) and a direction (east, northeast, etc.). Accelerating objects are either changing the magnitude of the velocity by speeding up or slowing down or changing the direction of the velocity by turning.

d. FALSE - Vice Versa. The acceleration of an object is inversely dependent upon the mass and directly dependent upon the net force.

e. FALSE - Acceleration is directly dependent upon the net force. Whatever alteration is made in the net force, the same alteration must be made in the acceleration. So if the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2 from 8 m/s/s to 16 m/s/s.

f. FALSE - Whatever alteration is made in the net force, the same alteration must be made in the acceleration. So if the net force is increased by a factor of 3, then the acceleration is increased by a factor of 3 from 8 m/s/s to 24 m/s/s.

g. FALSE - Acceleration is inversely dependent upon the mass. Whatever alteration is made in the mass, the inverse must be made of the acceleration. So if the mass is increased by a factor of 2, then the acceleration is decreased by a factor of 2 from 8 m/s/s to 4 m/s/s.

h. TRUE - Acceleration is inversely dependent upon the mass. Whatever alteration is made in the mass, the inverse must be made of the acceleration. So if the mass is increased by a factor of 4, then the acceleration is decreased by a factor of 4 from 8 m/s/s to 2 m/s/s.

i. FALSE - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2. If the mass is decreased by a factor of 2, then the acceleration is

increased by a factor of 2. The overall result of the two changes is to increase acceleration by a factor of 4 from 8 m/s/s to 32 m/s/s.

j. TRUE - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 2, then the acceleration is increased by a factor of 2. If the mass is decreased by a factor of 4, then the acceleration is decreased by a factor of 4. The overall result of the two changes is to decrease acceleration by a factor of 2 from 8 m/s/s to 4 m/s/s.

k. TRUE - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is decreased by a factor of 2, then the acceleration is decreased by a factor of 2. If the mass is decreased by a factor of 4, then the acceleration is decreased by a factor of 4. The overall result of the two changes is to decrease acceleration by a factor of 8 from 8 m/s/s to 1 m/s/s.

l. TRUE - Acceleration is inversely dependent upon the mass and directly dependent upon the net force. If the net force is increased by a factor of 4, then the acceleration is increased by a factor of 4. If the mass is increased by a factor of 2, then the acceleration is decreased by a factor of 2. The overall result of the two changes is to increase acceleration by a factor of 2 from 8 m/s/s to 16 m/s/s.

m. FALSE - The net force is the product m•a. Acceleration (a) can be calculated as the velocity change per time. The velocity change is +6 m/s (from 0 m/s to 6 m/s), so the acceleration is (+6 m/s) / (3 s) = +2 m/s/s. Therefore the net force is (2 kg)•(+2 m/s/s) = +4 N. The + indicates information about the direction; the 4 N is the magnitude.

n. FALSE - The net force is the product m•a. Acceleration (a) can be calculated as the velocity change per time. The velocity change is -15 m/s (from 24 m/s to 9 m/s), so the acceleration is (-15 m/s) / (3 s) = -5 m/s/s. Therefore the net force is (10 kg)•(-5 m/s/s) = -50 N. The - indicates information about the direction; the 50 N is the magnitude.

7. Big Bubba has a mass of 100 kg on the earth. What is Big Bubba's mass on the moon where the force of gravity is approximately 2/5-th that of Earth's? ________ Explain or show your work.

Answer: 100 kg

Mass is a quantity which is independent of the location of the object. So if Big Bubba has a mass of 100 kg on Earth, then he also has a mass of 100 kg on Mars. Only the weight would change as Big Bubba is moved from the Earth to Mars. He weighs ~1000 N on Earth and 1/6-th this value (~400 N) on the moon.

8. Little Billie weighs 360 N on Earth. What is Little Billie's mass on the moon where the force of gravity is approximately 1/6-th that of Earth's? ________ Explain or show your work.

Answer: ~36 kg

The mass of an object is related to weight by the equation W = m•g where g = ~10 m/s/s on Earth and one-sixth this value (~1.67) on the moon. So if Billy weighs 360 N on Earth, then his mass is approximately ~36 kg. His mass on the moon will be the same as his mass on Earth. Only his weight changes when on the moon; rather than being 360 N, it is 60 N. His weight on the moon could be found by multiplying his mass by the value of g on the moon: (36 kg) • (1.67 m/s/s) = ~60 N

9. TRUE or FALSE:

For an object resting upon a non-accelerating surface, the normal force is equal to the weight of the object.

Answer: False

Quite surprisingly to many, the normal force is not necessarily always equal to the weight of an object. Suppose that a person weighs 800 N and sits at rest upon a table. Then suppose another person comes along and pushes downwards upon the persons shoulders, applying a downward force of 200 N. With a total downward force of 200 N acting upon the person, the total upward force must be 1000 N. The normal force supplies the upward force to support both the force of gravity and the applied force acting upon the person. Its value is equal to 1000 N which is not the same as the force of gravity of the person.

 10. Which one(s) of the following force diagrams depict an object moving to the right with a constant speed? List all that apply.

 

Answer: AC

If an object is moving at a constant speed in a constant rightward direction, then the acceleration is zero and the net force must be zero. Choice B and E show a rightward net

force and therefore a rightward acceleration, inconsistent with the described motion.

11. As you sit in your chair and study your physics (presuming that you do), the force of gravity acts downward upon your body. The reaction force to the force of the Earth pulling you downward is ___.

a. the force of the chair pushing you upward

b. the force of the floor pushing your chair upward

c. the force of the Earth pushing you upward

d. the force of air molecules pushing you upwards

e. the force of your body pulling the Earth upwards

f. ... nonsense! Gravity is a field force and there is no such reaction force.

Answer: E

The most common wrong answer is a - the force of the chair pushing you upward. As you sit in your chair, the chair is indeed pushing you upward but this is not the reaction force to the force of the Earth pulling you downward. The chair pushing you upward is the reaction force to you sitting on it and pushing the chair downward. To determine the action-reaction force pairs if given a statement of the form object A pulls X-ward on object B, simply take the subject and the object in the sentence and switch their places and then change the direction to the opposite direction (so the reaction force is object B pulls object A in the opposite direction of X). So if the Earth pulls you downward, then you pull the Earth upward.

12. A golf pro places a ball at rest on the tee, lines up his shot, draws back his club, and lets one rip. During the contact of the golf club with the golf ball, the force of the club on the ball is ____ the force of the ball on the club and the acceleration of the club is ____ than the acceleration of the ball.

a. greater than, greater than b. greater than, equal to c. greater than, less than

d. less than, less than e. less than, equal to f. less than, greater than

g. equal to, equal to h. equal to, greater than i. equal to, less than

Answer: I

For every action, there is an equal and opposite reaction force. In this case, the force on the club is equal to the force on the ball. The subsequent accelerations of the interacting objects will be inversely dependent upon mass. The more massive club will have less acceleration than the less massive ball.

 

Each one of Newton's Laws can play a role in any one particular situation. However, one of the laws is often most obviously dominant in governing the motion of a situation. Pick which of Newton's most governs the situations described below.

a. First Law (inertia) b. Second Law (F = m•a) c. Third Law (action-reaction)

13. A helicopter must have two sets of blades in order to fly with stability.

14. If you were in an elevator and the cable broke, jumping up just before the elevator hit the ground would not save you. Sorry.

15. You usually jerk a paper towel from a roll in order to tear it instead of pulling it smoothly.

16. A student desk changes the amount of force it puts on other objects throughout a school day.

17. Heavy objects are not easier to move around in a horizontal fashion on the Moon than on the Earth.

18. The stronger, heavier team in a tug-of-war does not create a larger tension in the rope than the weaker, lighter team.

Answers: See answers and explanations below.

13. C - As the helicopter blades spin and push air in one direction, the air pushes the blades in the opposite direction; the result is that the helicopter can begin to rotate about the axis of the blade. To counteract this rotation, a second set of blades is required.

14. A - An object moving downwards will continue to move downwards unless acted upon by an unbalanced force. If you make an effort to supply such a force in an attempt to suddenly alter the direction of your motion, then you are creating a greater velocity change than if you merely hit the ground and stopped. If this greater velocity change occurred suddenly (in the same amount of time as the stopping of you and the elevator), then you

would experience a greater acceleration, a greater net force, and a greater ouch mark than if you had merely hit the ground and stopped.

15. A - The paper towel is at rest and resists changes in its at rest state. So if you apply a sudden force to one of the paper towel sheets, the great mass of the remainder of the roll will resist a change in its at rest state and the roll will easily break at the perforation.

16. C - As a student sits in the seat, they are applying a downward force upon the seat. The reaction force is that the seat applies an upward force upon the person. A weightier person will apply more downward force than a lighter person. Thus, the seat will constantly be changing the amount of reaction force throughout the day as students of different weight sit in it.

17. A - All objects have inertia or a tendency to resist changes in their state of motion. This inertia is dependent solely upon mass and is subsequently not altered by changes in the gravitational environment. To move an object horizontally, one must apply a force; this force will be resisted by the mass or inertia of the object. On the moon, the object offers the same amount of inertia as on Earth; it is just as difficult (or easy) to move around.

18. C - A rope encounters tension when pulled on at both ends. The tension in the rope is everywhere the same. If team A were to pull at the left end, then the left end would pull back with the same amount of force upon team A. This force is the same everywhere in the rope, including at the end where team B is pulling. Thus team B is pulling back on the rope with the same force as team A. So if the forces are the same at each end, then how can a team ever win a tug-of-war. The way a stronger team wins a tug-of-war is with their legs. They push upon the ground with a greater force than the other team. This force upon the ground results in a force back upon the team in order for them to pull the rope and the other team backwards across the line.

19. When a horse pulls a wagon, the force that causes the horse to move forward is….e. the force he exerts on the ground.f. the force he exerts on the wagon.g. the force the ground exerts on himh. the force the wagon exerts on him.

Answer: c) this is the only force acting on him in the forward direction

For the next several questions, consider the velocity-time plot below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter. 

20 During which time interval(s), if any, are there no forces acting upon the object? List all that apply.

21 During which time interval(s), if any, are the forces acting upon the object balanced.? List all that apply.

22 During which time interval(s), if any, is there a net force acting upon the object? List all that apply.

23 During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

24 During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.

Answers: See answers and explanations below.

20 None - If an object is on a surface, one can be guaranteed of at least two forces - gravity and normal force.

21 BDFH - If the forces are balanced, then an object is moving with a constant velocity. This is represented by a horizontal line on a velocity-time plot.

22 ACEG - If an object has a net force upon it, then it is accelerating. Acceleration is represented by a sloped line on a velocity-time plot.

23 AE - If the net force is directed to the right, then the acceleration is to the right (in the + direction). This is represented by a line with a + slope (i.e., upward slope).

24 CG - If the net force is directed to the left, then the acceleration is to the left (in the - direction). This is represented by a line with a - slope (i.e., downward slope).

For the next several questions, consider the dot diagram below for the motion of an object along a horizontal surface. The motion is divided into several time intervals, each labeled with a letter.

25. During which time interval(s), if any, are there no forces acting upon the object? List all that apply.

26. During which time interval(s), if any, are the forces acting upon the object balanced.? List all that apply.

27. During which time interval(s), if any, is there a net force acting upon the object? List all that apply.

28. During which time interval(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

29. During which time interval(s), if any, is the net force acting upon the object directed toward the left? List all that apply.

Answers: See answers and explanations below.

25. None - If an object is on a surface, one can be guaranteed of at least two forces - gravity and normal force.

26. ACEGI - If the forces are balanced, then an object is moving with a constant velocity or at rest. This is represented by a section of a dot diagram where the dots are equally spaced apart (moving with a constant velocity) or not even spaced apart at all (at rest).

27. BDFH - If an object has a net force upon it, then it is accelerating. Acceleration is represented by a section of a dot diagram in which the spacing between consecutive dots is either increasing or decreasing.

28. BF - If the net force is directed to the right, then the acceleration is to the right (in the + direction). This is represented by a dot diagram in which the dots are increasing their separation distance as the object moves from left to right.

29. DH - If the net force is directed to the left, then the acceleration is to the left (in the - direction). This is represented by a dot diagram in which the dots are decreasing their separation distance as the object moves from left to right.

 

For the next several questions, consider the trajectory diagram shown below for a projectile thrown at an angle to the horizontal. The vector arrows represent the horizontal and vertical components of the projectile's velocity. Several points in the trajectory are labeled with a letter. Use the trajectory diagram to answer the next several questions. (Consider air resistance to be negligible.)

30. At which point(s), if any, are there no forces acting upon the object? List all that apply.

31. At which point(s), if any, are the forces acting upon the object balanced.? List all that apply.

32. At which point(s), if any, is there a net force acting upon the object? List all that apply.

33. At which point(s), if any, is the net force acting upon the object directed toward the right? List all that apply.

34. At which point(s), if any, is the net force acting upon the object directed upward? List all that apply.

Answers: See answers and explanations below

30. None

31. None

32. ABCDEFG

33. None

34. None

This object is a projectile as can be seen by its constant horizontal velocity and a changing vertical velocity (besides that, the problem states that this is a projectile). A projectile is an object upon which the only force is gravity. Gravity acts downward to accelerate an object downward. This force is an unbalanced force or net force. It causes a vertical rising object to slow down and a falling object to speed up.

The presence of a horizontal velocity does not demand a horizontal force, only a balance of horizontal forces. Having no forces horizontally would cause the projectile to move at a constant horizontal speed once it is launched. Similarly an upward force is not needed on this projectile. When launched, an upward velocity is imparted to it; this velocity is steadily decreased as the object is acted upon by the downward force of gravity. An upward force would only be required for an object which is speeding up as it rises upward.

35. Construct free-body diagrams for the following physical situations at the instant in time for which they are described. As is always done in free-body diagrams, label the forces according to type and draw the arrows such that their length reflectshe magnitude of the force.

a. A book is being pushed to the right across a table surface with a constant velocity. (Neglect Fair.)

b. A rightward-moving box (which was previously set into rightward motion across the floor) gradually slows to a stop.

 

c. A falling skydiver has reached a terminal velocity.

 

d. A student is pushing lightly upon a large box in an attempt to push it to the right across the floor, but the box fails to move.

 

e. An elevator (held by a cable) is moving upwards at a constant velocity. (Neglect Fair.)

f. A picture hangs symmetrically by two wires oriented at angles to the vertical.

Answer:  

a. A book is being pushed to the right across a table surface with a constant velocity. (Neglect Fair.)

 

b. A rightward-moving box (which was previously set into rightward motion across the floor) gradually slows to a stop.

c. A falling skydiver has reached a terminal velocity.

d. A student is pushing lightly upon a large box in an attempt to push it to the right across the floor, but the box fails to move.

e. An elevator (held by a cable) is moving upwards at a constant velocity. (Neglect Fair.)

f. A picture hangs symmetrically by two wires oriented at angles to the vertical.

 

For the following problems, draw free-body diagrams and solve for the requested unknown. Use g = 9.8 m/s/s.

36. A 72-kg skydiver is falling from 10 000 feet. After reaching terminal velocity, the skydiver opens his parachute. Shortly thereafter, there is an an instant in time in which the skydiver encounters an air resistance force of 1180 Newtons. Determine the acceleration of the skydiver at this instant. PSYW

Answer: 6.6 m/s/s, up

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. The sum of the vertical forces is

∑Fy = 1180 N, up + 706 N, down = 474 N, up

The acceleration of the skydiver can be computed using the equation ∑Fy = m•ay.

ay = (474 N, up) / (72 kg) = 6.59 m/s/s, up

37. A 5.2-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine the acceleration of the object. (Neglect air resistance.) PSYW

Answer: 5.0 m/s/s, right

Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 5.2 N, right (equal to the only rightward force - the applied

force). So the acceleration of the object can be computed using Newton's second law.

 a = Fnet / m = (5.2 N, right) / (1.05 kg) = 4.95 m/s/s, right

38. A 921-kg sports car is moving rightward with a speed of 29 m/s. The driver suddenly slams on the brakes and the car skids to a stop over the course of 3.2 seconds with the wheels locked. Determine the average resistive force acting upon the car. PSYW

Answer: 8.3 x 103 N

There are three (perhaps four) forces acting upon this car. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding car. In the free-body diagram, these two forces are represented by the Ffrict arrow.

The acceleration is not given but can be calculated from the kinematic information that is given:

vi = 29 m/s, vf = 0 m/s, and t = 3.2 s

The acceleration of the object is the velocity change per time:

a = Delta v / t = (0 m/s - 29 m/s) / (3.2 s) = -9.67 m/s/s or 9.67 m/s/s, left.

This acceleration can be used to determine the net force:

Fnet = m•a = (921 kg) • (9.67 m/s/s, left) = 8347 N, left

The friction forces (surface and air) provide this net force and are equal in magnitude to this net force.

39. A tow truck exerts a 18300-N force upon a 1200-kg car to drag it out of a mud puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane of motion of the car is horizontal. Determine the time required to drag the car a distance of 6.9 meters from its rest position. PSYW

Answer: 6.4 s

Upon neglecting air resistance, there are four forces acting upon the object. The up and down forces balance each other. The acceleration is rightward (or in the direction of the applied force) since the rightward applied force is greater than the leftward friction force. The horizontal forces can be summed as vectors in order to determine the net force.

Fnet = ·Fx = 18300 N, right - 17900 N, left = 400 N, right

The acceleration of the object can be computed using Newton's second law.

 ax = ·Fx / m = (400 N, down) / (1200 kg) = 0.333 m/s/s, right

This acceleration value can be combined with other known kinematic information (vi = 0 m/s, d = 6.9 m) to determine the time required to drag the car a distance of 6.9 m. The following kinematic equation is used; substitution and algebra steps are shown.

d = vi • t + 0.5 •a • t2

d = vi • t + 0.5 •a • t2

6.9 m = 0.5 • (0.333 m/s/s) • t2

6.9 m / (0.5 • 0.333 m/s/s ) = t2

41.4 = t2

6.43 s = t

40. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the speed (in m/s) of the bucket after 1.59 seconds. PSYW

Answer: 14.2 m/s

There are two forces acting upon the bucket - the force of gravity (up) and the tension force (down). The magnitude of the force of gravity is found from m•g; its value is 43.5 N. These two forces can be summed as vectors to determine the net force.

Fnet = ·Fy = 83.1 N, up + 43.5 N, down = 39.6 N, up

The acceleration can be calculated using Newton's second law of motion.

a = Fnet / m = (39.6 N, up) / (4.44 kg) = 8.92 m/s/s, up

The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 1.59 s) to calculate the final speed of the bucket. The kinematic equation, substitution and algebra steps are shown.

vf = vi + a•t

vf = 0 m/s + (8.92 m/s/s)•(1.59 s)

vf = 14.2 m/s

41. A 22.6-N horizontal force is applied to a 0.071-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds.

Answer: 23 m/s

Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.

 a = Fnet / m = (22.6 N, right) / (0.071 kg) = 318 m/s/s, right

The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown.

vf = vi + a•t

vf = 0 m/s + (318 m/s/s)•(0.0721 s)

vf = 23.0 m/s

42. An object rests upon an inclined plane. If the angle of incline is increased, then the normal force would _______.

a. increase b. decrease c. remain the sameAnswer: B

The normal force is equal to the perpendicular component of the weight vector. So Fnorm = Fperp = m•g•cos(angle). If the incline angle is increased, the cos(angle) decreases towards 0; thus, the normal force decreases as well. Of course, the extreme is when the angle is 90 degrees and there is no normal force.

43. Three pictures of equal weight (20 N) are hung by wires in three different orientations. In which orientation are the wires least likely to break? _______ Explain why.

  Answer: 60-degree angle to the horizontal

The tension in the wire will have the greatest impact upon its tendency to break. As the wire becomes most vertically oriented, the horizontal component of the tension force is reduced and the tension becomes less. The moral of the story -- to support the weight of a picture, one only needs to pull upwards, not leftwards and rightwards. (Of course, the best arrangement would be two wires pulling completely vertically.)

44. A 50-N force is applied at an angle of 30 degrees north of east. This would be the same as applying two forces at

e. 43 N, east and 7 N, north f. 35 N, east and 15 N, north g. 25 N, east and 25 N, north h. 43 N, east and 25 N, north

Answer: D

A 50-N force at 30 degrees would have two components - 43 N, east and 25 N, north. These two forces would be equal to the 50-N force at 30 degrees N of E.

Use the approximation that g=~10 m/s2 to fill in the blanks in the following diagrams. DON’T WORRY ABOUT SIG FIGS FOR QUES 45 - 51

45.

 

46.

 

 

47.

 

48.

 

49.

 

50.

 51.

 

 

45.

 Fgrav = m•g = 800 N

∑Fy = may = (80 kg)•(2.0 m/s/s)

∑Fy = 160 N, down

The Fgrav (down) and the Fair (up) must add up to 160 N, down. Thus, Fair must be smaller than Fgrav by 160 N.

Fair = 640 N

46.

 

   Fgrav = m•g = 800 N

Since there are two forces pulling upwards and since the sign is hanging symmetrically, each force must supply an upwards pull equal to one-half the object's weight. So the vert

pull (Fy) in each force is 400 N. The following triangle can be set up:

Using trig, we can write:

sin(30 deg.)=(400 N)/Ftens

Solving for Ftens yields 800 N.

47.

 

A quick blank is Fgrav:  Fgrav = m•g = 800 N

Now resolve the 60-N force into components using trigonometry and the given angle measure:

Fx = 60 N•cos(30 deg) = 52 N

Fy = 60 N•sin(30 deg) = 25 N

Since the acceleration is horizontal, the sum of the vertical forces must equal 0 N. So Fgrav

= Fy + Fnorm.

Therefore Fnorm = Fgrav - Fy = 50 N.

Knowing Fnorm and mu, the Ffrict can be determined:

Ffrict = mu•Fnorm = 0.5*(50 N) = 25 N

Now the horizontal forces can be summed:

∑Fx = Fx + Ffrict = 52 N, right + 25 N, left

∑Fx = 27 N, right

Using Newton's second law, ∑Fx = m•ax

So ax = (27 N)/(8 kg) = 3.8 m/s/s, right

48.

 

The first step in an inclined plane problem is to resolve the weight vector into parallel and perpendicular components:

Fpar = m•g•sin(angle) = (420 N)•sin(30 deg) = 210 N

Fperp = m•g•cos(angle) = (420 N)•cos(30 deg) = 364 N

The mass can be found as m = Fgrav/g

m = Fgrav/g = (420 N)/(10 m/s/s) = 42 kg

The Fnorm acts opposite of and balances the Fperp.

So Fnorm = Fperp = 364 N

Knowing Fnorm and mu, the Ffrict can be determined:

Ffrict = mu•Fnorm = 0.2*(364 N) = 73 N

Now the forces parallel to the incline can be summed:

∑F|| = F|| + Ffrict = 210, down to left + 73 N, up to right

∑F|| = 137 N, down to left

Using Newton's second law, ∑F|| = m•a||

So a|| = (137 N)/(42 kg) = 3.3 m/s/s

49.

 

Treating the two masses as a single system, it can be concluded that the net force on the 9-kg system is:

∑Fsystem = m•asystem = (9 kg)•(2.5 m/s/s) = 22.5 N, right

The free-body diagram for the system is:

The Fnorm supporting the 9-kg system is ~90 N.

So the Ffrict acting upon the system is:

Ffrict = mu• Fnorm = 0.20*(90 N) = 18 N, left

So if ∑Fsystem = 22.5 N, right and Ffrict = 18 N, left, the rightward Ftens1 must equal 40.5 N.

The Ftens2 force is found inside the system; as such it can not be determined through a system analysis. To determine the Ftens2, one of the masses must be isolated and a free-

body analysis must be conducted.

The 3-kg mass is selected and analyzed:

The Fnorm and Fgrav balance each other; their value is ~30 N. The Ffrict on the 3-kg mass is:

Ffrict = mu•Fnorm = 0.20*(30 N) = 6 N, left

The net force on the 3-kg object is:

∑Fx = m•ax = (3 kg)•(2.5 m/s/s) = 7.5 N, right

The horizontal forces must sum up to the net force on the 3-kg object; So

∑Fx = m•ax = Ftens2 (right) + Ffrict (left)

7.5 N, right = Ftens2 + 6 N, left

The Ftens2 must be 13.5 N.

50.

 

 Like most two-body problems involving pulleys, it is usually easiest to forgo the system analysis and conduct separate free-body analyses on the individual masses. Free-body

diagrams, the chosen axes systems, and associated information is shown below.

Analyzing the Fx forces on the 250-g mass yields:

max = Ftens - Ffrict

Since Ffrict = mu•Fnorm and Fnorm = 2.5 N

The Ffrict is (0.1)*(2.5 N) = 0.25 N.

Substituting into equation 1 yields

(0.250 kg)•ax = Ftens - 0.25 N

 

Analyzing the Fy forces on the 50-g mass yields:

may = Fgrav - Ftens

Substituting m and Fgrav values into equation 3 yeilds:

(0.050 kg)*ay = (0.500 N) - Ftens

The above equation can be rearranged to:

Ftens = (0.500 N) - (0.050 kg)*ay

Equation 4 provides an expression for Ftens; this can be substituted into equation 2:

(0.250 kg)•ax = (0.500 N) - (0.050 kg)*ay - 0.25 N

Now since both masses accelerate at the same rate, ax =ay

and the above equation can be simplified into an equation with 1 unknown - the acceleration (a):

(0.250 kg)•a = (0.500 N) - (0.050 kg)*a - 0.25 N

After a few algebra steps, the acceleration can be found:

(0.0300 kg)•a = 0.25 N

a = 0.833 m/s/s

Now that a has been found, its value can be substituted back into equation 4 in order to solve for Ftens:

Ftens = (0.500 N) - (0.050 kg)*(0.833 m/s/s)

Ftens = 0.458 N

51.

 

This problem can most easily be solved using separate free-body analyses on the individual masses. Free-body diagrams, the chosen axes systems, and associated

information is shown below.

Note that the positive y-axis is chosen as being downards on the 200-g mass since that is the direction of its acceleration. Similarly, it chosen as upwards on the 100-g mass since

that is the direction of its acceleration.

For the 200-gram mass, the sum of the vertical forces equals the mass times the acceleration:

 Fgrav - Ftens = m•ax=y

2.00 N - Ftens = (0.200 kg)•ay

The same type of analysis can be conducted for the 100-gram mass:

Ftens - Fgrav = m•ay

Ftens - 1.00 N = (0.100 kg)•ay

Equation 2 can be rearranged to obtain an expression for the tension force:

Ftens = (0.100 kg)•ay + 1.00 N

This expression for Ftens can be substituted into equation 1 in order to obtain a single equation with acceleration (ay) as the unknown. The ay value can be solved for.

2.00 N -[(0.100 kg)•ay + 1.00 N] = (0.200 kg)•ay

2.00 N - 1.00 N = (0.200 kg)•ay + (0.100 kg)•ay

1.00 N = (0.300 kg)•ay

ay = (1.00 N)/(0.300 kg) = 3.33 m/s/s

Now with ay known, its value can be substituted into equation 3 in order to determine the tension force:

Ftens = (0.100 kg)•ay + 1.00 N

Ftens = (0.100 kg)•(3.33 m/s/s) + 1.00 N

Ftens = 0.333 N + 1.00 N = 1.33 N

52. A 945-kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of friction between tires and road is 0.972, determine the distance required to stop. PSYW

 

Answer: 26.8 m

Like most problems, this problem begins with a free-body diagram (as shown at right). Note that there is no rightwards applied force (a common mistake).

Note also that the force of friction is the only force responsible for the acceleration (deceleration) of the car. The Ffrict value is the net force. So determining the acceleration involves finding Fgrav (Fgrav=m•g = 945•9.8 = 9261 N), Fnorm (the same as Fgrav= 9261 N), and Ffrict (Ffrict = mu•Fnorm = 0.972 • 9261 N = 9002 N). With the ·Fx = 9002 N, the acceleration can be calculated (ax = ·Fx/m = 9002 N/945 kg = 9.53 m/s/s.) This is a leftwards acceleration; ax will be assigned the numerical value of -9.53 m/s/s in the next part of this problem.

Now in the kinematic part of this problem, the distance must be found using the known information (vf = 0 m/s, vo = 22.6 m/s and a = 9.53 m/s/s). Use the equation:

vf2 = vo

2 + 2 a d

(0 m/s)2 = (22.6 m/s)2 + 2•(-9.53 m/s/s)•d

d = [(22.6 m/s)2] / [2•(9.53 m/s/s)] = 26.8 m

53. A student pulls a 2.00-kg backpack across the ice (assume frictionless) by pulling at a 30.0° angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force. PSYW

 

Answer: 0.289 N

The acceleration of the object can be found from the slope of the line on the v-t graph. The acceleration is:

a = change in velocity/change in time = (2.0 m/s) / 16 s = 0.125 m/s/s.

The net force is found from m•a; so Fnet = (2 kg)•(0.125 m/s/s) = 0.250 N.

The only force that can contribute to the net force is the applied force. The horizontal component of the applied force must therefore be equal to 0.250 N. A diagram depicting the relationship of the lengths of the side for this 30-60-90 triangle is shown at the right. The cosine function can be utilized to determine the magnitude of the tension force.

cos(30 deg) = 0.250 N)/Fapp

Solving for Fapp yields 0.289 N.

54. Splash Mountain at Disney World in Orlando, Florida is the steepest water plume ride in the country. Occupants of the boat fall from a height of 100.0 feet (3.2 ft = 1 m) down a wet ramp which makes a 45° angle with the horizontal. Consider the mass of the boat and its occupants to be 1500 kg. The coefficient of friction between the boat and the ramp is 0.10. Determine the frictional force, the acceleration, the distance traveled along the incline, and the final velocity of the boat at the bottom of the incline. PSYW

Answer: Ffrict = 2.1 x 103 N; a = 5.5 m/s/s; d = 44 m; vf = 22 m/s

This is a multi-part inclined plane problem. Like all problems, it should begin with a free-body diagram. (If this is where the difficulty lies for you, then take some time to review inclined planes. See help page.) The friction force opposes the motion of

the boat. If the + x direction is defined as down and along the incline (as shown at the right) then the net force can be computed by the expression F|| - Ffrict. The parallel component of the weight vector (m•g•sin theta) is 10395 N. The Ffrict force is computed by multiplying the coefficient of friction (0.10) by the normal force (Fnorm). The normal force is balancing Fperp (m•g•cos theta) and is equal to it. So Fnorm = 10395 N and Ffrict = 2079 N. The net force (Fnet = F|| - Ffrict ) is 8316 N; and the acceleration is Fnet/m or 5.54 m/s/s.

Once the Newton's laws analysis has been completed and the acceleration determined, the kinematics portion of the problem can be tackled. First, determine the distance along the incline from the height of the hill and the angle of incline. The height of 100 ft is equivalent to 31.25 m; the distance along the incline is 44.19 m [found from (31.25 m)/sin (45)]. The final velocity is found using the acceleration, distance and a kinematic equation:

vf2 = vo

2 + 2•a•d

vf2 = (0 m/s)2 + 2•(5.54 m/s/s)•(31.25 m) = 489.98 m2/s2

vf = 22 m/s

55. At last year's Homecoming Pep Rally, Trudy U. Skool (attempting to generate a little excitement) slid down a 42° incline from the GBS dome to the courtyard below. The coefficient of friction between Trudy's jeans and the incline was 0.650. Determine Trudy's acceleration along the incline. Begin with a free-body diagram. PSYW

  Answer: 1.8 m/s2

This is nearly an identical force analysis as the last problem except the mass is not known. When the mass is not known, the problem-solving strategy involves inserting m into the equation as an unknown variable and proceding with the solution. It is likely that in a subsequent step of the problem that the m will cancel and the accceleration can be determined without knowing m.

The net force is found by the expression F|| - Ffrict. The parallel component of the weight vector is m•g•sin(theta). The Ffrict force is computed by multiplying the coefficient of friction (mu) by the normal force (Fnorm). The normal force is balancing Fperp and is equal to m•g•cos(theta). So Fnorm = m•g•cos(theta) and Ffrict = mu•m•g•cos(theta). The net force is m•g•sin(theta) - mu•m•g•cos(theta). The acceleration is Fnet/m or [m•g•sin(theta) - mu•m•g•cos(theta)]/m. Note that there is an m in both terms in the numerator and a m in the denominator. The masses cancel (I love that part) and the equation reduces to

a = [g•sin(theta) - mu•g•cos(theta)]

By substitution of the given values into the equation, it can be shown that a = 1.82 m/s/s.

56. Consider the two-body system at the right. There is a 0.250-kg object accelerating across a rough surface. The sliding object is attached by a string to a 0.10 kg object which is suspended over a pulley. The coefficient of kinetic friction is 0.183. Calculate the acceleration of the block and the tension in the string. PSYW

 

Answer: a = 1.5 m/s2 ; Ftens = 0.83 N

The solution to this problem begins by drawing a free-body diagram for each object. Note that the + x-axis for the 0.25-kg object is drawn in the direction that the object accelerates; and the + y-axis for the 0.10-kg object is drawn in the direction which it accelerates.

Note that while friction acts upon the 0.25-kg object, it does not act upon the 0.10-kg object (since it is not being dragged across the surface). Newton's second law of motion (·F = m•a) can be applied to the motion of the 0.25-kg mass:

∑Fx = m•ax

Ftens - Ffrict = m•ax

The expressioin for Ffrict can be substituted into the equation.

Ftens - mu•Fnorm = m•ax

Since the vertical forces on the 0.25-kg object balance each other, it is known that Fgrav = Fnorm; so Fnorm = m•g = (0.25 kg)•(9.8 m/s/s) = 2.45 N. Since mu = 0.183, mu•Fnorm = 0.448 N. Equation 1 can now be re-written as

Ftens - 0.448 N = (0.25 kg)•ax

Newton's second law of motion (·F = m•a) can be applied to the motion of the 0.10-kg mass:

∑Fy = m•ay

Fgrav - Ftens = m•ay

The expression for Fgrav (m•g) can be substituted into the equation.

m•g - Ftens = m•ay

(0.10 kg)•(9.8 m/s/s) - Ftens = m•ay

0.98 N - Ftens = m•ay

The mass of the 0.10-kg object can be substituted into the equation to yield equation 3.

0.98 N - Ftens = (0.10 kg)•ay

Equations 2 and 3 each include an acceleration term - ax and ay. These acceleration values are the same since the system accelerates together; thus, ax = ay = a. An inspection of these two equations show that there are now two equations and two unknowns - Ftens and a. These unknown values can be solved for in the customary manner.

First, equation 3 is rearranged to create an expression for Ftens:

0.98 N - (0.10 kg)•a = Ftens

This expression for Ftens is now substituted into equation 2.

Ftens - 0.448 N = (0.25 kg)•a

0.98 N - (0.10 kg)•a - 0.448 N = (0.25 kg)•a

This equation can now be solved for acceleration (a) as shown in the following steps.

0.98 N - 0.448 N = (0.25 kg)•a + (0.10 kg)•a

0.532 N = (0.35 kg)•a

a = 1.52 m/s/s

Now that a is known, its value can be substituted into equation 4 in order to solve for Ftens.

0.98 N - (0.10 kg)•a = Ftens

0.98 N - (0.10 kg)•(1.52 m/s/s) = Ftens

0.98 N - 0. 152 N = Ftens

0.83 N = Ftens

57.

Answer: 11 m

58. Two boxes resting on a frictionless surface are connected by a massless cord. The boxes have masses of 14.0 and 11.0 kg. A horizontal force of Fapp = 50.0 N is applied by pulling on the lighter box.What is the acceleration of each box and the tension in the cord?

i. Draw a freebody diagram for each box:

ii. Write equations for the free body diagrams for box 1 and box 2:

Box 1 (lighter): Fnet1 = Fapp – FT

Box 2 (heavier): Fnet2 = FT

iii. The boxes must have identical acceleration a1=a2 otherwise the cord would part or bunch up which is counter to our experience of tension between the boxes.

In order to find the acceleration, we must add the two equations:

Fnet1 = Fapp – Fnet2

m1a = Fapp – m2aFapp = a(m1 + m2)a = 50.0/25.0 = 2.00 m/s2

To determine tension, substitute m2 into box 2 equation (could use box 1 equation as well): Remember that the force of tension is the pull of box 2 on box 1, therefore it is dependent on the mass at the end of the string.

FT = m2a = 14.0 * 2.00 = 22.0 N

It makes sense for the tension to be less than the pulling force since the tension accelerates only the second box.

Note how analyzing two free-body diagrams allows us to understand an “internal force” like Tension in the rope. Summed together, tension would make zero contribution to the net force on the whole system.