Ch6.1 - Forces Newton's 1st Law - object in motion will remain in that motion, an object at...

Ch6.1 - ForcesNewton's 1st Law - object in motion will remain in that motion,

an object at rest stays at rest.

- Will not change unless a net force acts on it.

- Rest is just a special case of motion.

Newton's 1st Law - object in motion will remain in that motion, an object at rest stays at rest.

Inertia - tendency of an object to resist change.

Ch6.1 - Forces

Newton's 1st Law - object in motion will remain in that motion, an object at rest stays at rest.

Inertia - tendency of an object to resist change.

Force - push or pull

Contact force - touches object

Field force - long range, acts at a distance

Equilibrium - net force on an object is zero(object can be at rest or moving at constant speed)

Free body diagram - drawing of object with all forces labeled on it.

Ch6.1 - Forces

Net Force - what’s left over after all forces we added up. Force is a vector, so the vectors have to be added vectorally!

Ex1) What is the net force in each:

Ch6.1 - Forces

Net Force - what’s left over after all forces we added up. Force is a vector, so the vectors have to be added vectorally!

Ex1) What is the net force in each:

Ch6.1 - Forces

Newton’s 2nd Law:F = (m)(a)

Units of force: = Newtons

Fa net

Units of force: = Newtons

Types of Forces:

Fg = Gravity FN = Normal force FT = Tension (weight)

Don’t confuse Fnet with FN

Fa net

Ex 2) A rope lifts a bucket, its speed is increasing. Draw and label forces. What is net force?

Fa net

HW#1) Draw free body diagrams for:a. A book held stationary in hand.

b. Book pushed across a table at constant speed.

HW #7) A skydiver falls downward thru the air, accelerating

HW #4) A 225N force is exerted on a crate to the North. A 165N force is also exerted on the crate, but east. Find the net force, both magnitude and direction.

CH 6 HW #1: 1-8

Lab6.1 – Inertia

- due tomorrow

- Ch6 HW#1 due at beginning of period

Ch6 HW#1 1- 81. Draw: c. A book on the desk while you hand is

pushing down on it.d. A ball, just as the string holding it breaks.

2. 2 horiz forces 225N and 165N pushing on crate in same direction.Find Fnet.

3. 2 horiz forces 225N and 165N pushing on crate in opp directions.Find Fnet.

Ch6 HW#1 1- 8 FN

1. Draw: c. A book on the desk while you hand is pushing down on it.d. A ball, just as the string holding it breaks. Fg

c. Fnet = Fg + FHand – FN d. Fnet = Fg 0 = Fg

2. 2 horiz forces 225N and 165N pushing on crate in same direction.Find Fnet.

Ch6 HW#1 1- 8 FN

2. 2 horiz forces 225N and 165N pushing on crate in same direction.Find Fnet. FN

F = 165N F = 225N

Fnet = 165N + 225N = 390N Fg

Ch6 HW#1 1- 8 FN

2. 2 horiz forces 225N and 165N pushing on crate in same direction.Find Fnet. FN

F = 165N F = 225N

Fnet = 165N + 225N = 390N Fg

3. 2 horiz forces 225N and 165N pushing on crate in opp directions.Find Fnet. FN

F = 165N F = 225N

Fnet = 225N – 165N = 60N to the right Fg

5. Calc the force you exert on the floor while you stand. (1 lb = .454kg)Does the force change if you lie down?

6. A cable pulls a crate at constant speed across the floor.

8. A rope lowers a bucket at constant speed.

5. Calc the force you exert on the floor while you stand. (1 lb = .454kg)Does the force change if you lie down? FN

150 lb .454 kg = 68 kg x 9.8 m/s2 1 lb

= 680 N

6. A cable pulls a crate at constant speed across the floor.

150 lb .454 kg = 68 kg x 9.8 m/s2 1 lb

= 680 N

6. A cable pulls a crate at constant speed across the floor. FN Fnet = Fg – FN

0 = Fg – FN Ff FT Fg = FN

150 lb .454 kg = 68 kg x 9.8 m/s2 1 lb

= 680 N

6. A cable pulls a crate at constant speed across the floor. FN Fnet = Fg – FN

0 = Fg – FN Ff FT Fg = FN

8. A rope lowers a bucket at constant speed. FT

Fnet = Fg – FT 0 = Fg – FT Fg = FT

Ch6.2 – Mass vs. Weight

-All objects in Free Fall accelerate at g= 9.8m/s

F = m.a

Ch6.2 – Mass vs. Weight

-All objects in Free Fall accelerate at g= 9.8m/s

F = m.a

Force of gravity: Fg= m.g

Weight – measure of the force of gravity (Newtons, N)

Mass – amount of matter present (kilograms, kg.)

Ex1) My mass is 80 kg, a. What’s my weight? b. What’s my mass on the moon? c. If gravity on the moon is 1/6g, what’s my weight on the moon?

If you are accelerating up or down (like in an elevator), your weight changes.

- If it accelerates upward, you squish the scale more & you appear to weigh more.

F = m(g + a)

F = m(g + a) - If it accelerates downward, the scale expands

& you apparently weigh less. F = m(g – a)

& you apparently weigh less. F = m(g – a) Ex2) Your mass is 75kg, you stand on a scale

in an elevator not moving. a. What is your weight? b. If the elevator accelerates UP at 2 m/s

2 , what's your weight?

c. If it is moving at a constant speed what's your weight?

d. Accelerates down at 2 m/s2?

e. Accelerates down at 5 m/s2?

f. Accelerates down at 9.8 m/s2?

& you apparently weigh less. F = m(g – a) Ex2) Your mass is 75kg, you stand on a scale

in an elevator not moving. a. What is your weight? Fg= m(g+a)= (75kg)(9.8+0)= 735N b. If the elevator accelerates UP at 2 m/s

2 , what's your weight? Fg= m(g+a)= (75kg)(9.8+2)= 885N

c. If it is moving at a constant speed what's your weight? Fg= m(g+a)= (75kg)(9.8+0)= 735N

d. Accelerates down at 2 m/s2? Fg= m(g-a)= (75kg)(9.8-2)= 600N

e. Accelerates down at 5 m/s2? Fg= m(g-a)= (75kg)(9.8-5)= 375N

f. Accelerates down at 9.8 m/s2? Fg= m(g-a)= (75kg)(9.8-9.8)= 0N

Apparent weightlessness!

Ex3) A 50kg bucket is lifted by a rope that has a max tension of 550N. It started at rest, and after 3m is moving at 3 m/s, will the rope break?

Ch6 HW#2 9 – 13

Lab6.2 – Newton’s 2nd Law of Skateboards

- due tomorrow

Ch6 HW#2 9 – 13 9. Draw. A skydiver falls downward through the air

at a constant velocity.10. Draw. A rope lifts a bucket accelerating it upward.11. Draw. A rocket blasts off and its vertical velocity

increases with time. (No air drag.)12. You weigh 585N on Earth.

a. Mass?b. Weight on moon? (g=1.60 m/s

2)13. Mass of 60kg. What does scale read:

a. Elevator not moving?b. Up @ const speed?c. If the elevator were moving downward and then began slowing to a stop, scale reads ________.d. If the elevator were moving upward and then began slowing to a stop, scale reads ________.e. Slows at 2.0 m/s

2 while moving upward?f. Speeds up at 2.0 m/s

2 while moving downward?

Ch6 HW#2 9 – 13 Fair 9. Draw. A skydiver falls downward through the air

at a constant velocity.10. Draw. A rope lifts a bucket accelerating it upward.11. Draw. A rocket blasts off and its vertical velocity Fg

a. Mass?b. Weight on moon? (g=1.60 m/s

Ch6 HW#2 9 – 13 Fair FT

9. Draw. A skydiver falls downward through the air at a constant velocity.

10. Draw. A rope lifts a bucket accelerating it upward.11. Draw. A rocket blasts off and its vertical velocity Fg

a. Mass? Fg b. Weight on moon? (g=1.60 m/s

Ch6 HW#2 9 – 13 Fair FT Frocket

a. Mass? Fg Fg

b. Weight on moon? (g=1.60 m/s2)

13. Mass of 60kg. What does scale read:a. Elevator not moving?b. Up @ const speed?c. If the elevator were moving downward and then began slowing to a stop, scale reads ________.d. If the elevator were moving upward and then began slowing to a stop, scale reads ________.e. Slows at 2.0 m/s

a. Mass? 58.5kg Fg Fg

b. Weight on moon? (g=1.60 m/s2) Fgm = (58.5kg)(1.60 m/s

2) = 93.6N13. Mass of 60kg. What does scale read:

a. Elevator not moving? Fg= m(g+a)= (60kg)(10+0)= 600Nb. Up @ const speed? Fg= m(g+a)= (60kg)(10+0)= 600Nc. If the elevator were moving downward and then began slowing to a stop, scale reads _more_.d. If the elevator were moving upward and then began slowing to a stop, scale reads _less_.e. Slows at 2.0 m/s

2 while moving upward? f. Speeds up at 2.0 m/s

a. Elevator not moving? Fg= m(g+a)= (60kg)(10+0)= 600Nb. Up @ const speed? Fg= m(g+a)= (60kg)(10+0)= 600Nc. If the elevator were moving downward and then began slowing to a stop, scale reads _more_.d. If the elevator were moving upward and then began slowing to a stop, scale reads _less_.e. Slows at 2.0 m/s

2 while moving upward? Fg= m(g-a)= (60kg)(10-2) = 480Nf. Speeds up at 2.0 m/s

2 while moving downward? Fg= m(g-a)= (60kg)(10-2) = 480N

Ch6.3 – Friction Kinetic Friction- (Ff,k ) Moving Friction

Ff,k = µk × FN.

Coefficient Normal force of kinetic friction (usually equal & opposite to Fg)

Ex1) Calculate the kinetic friction for the steel blades of ice skates on ice given the total mass of the skater is 76 kg and µk =.06

Ch6.3 – Friction Kinetic Friction- (Ff,k ) Moving Friction

Ff,k = µk × FN.

Coefficient Normal force of kinetic friction (usually equal & opposite to Fg)

Ex1) Calculate the kinetic friction for the steel blades of ice skates on ice given the total mass of the skater is 76 kg. and µk =.06 (Fg = (75kg.)(10 m/s

2)= 750N)

FNFF,k= µk × FN

=(.06)Fg

=(.06)(750N)= 45N

Ex2) What is the coefficient of friction for steel on steel without lubrication, if the mass on the block on top is 10kg, and the friction force is measured to be 30N.

Ff,k = µk × FN

30N = µk(100N)

µk = .3

Static Friction- (Ff,s) Not moving friction (prevents motion from happening)

Ff,s = µs × FN.

Coefficient Normal force of static friction

Ff,s = µs × FN.

Static Friction can vary between 0 and this maximum value

Ex3) A person pushes on a wooden crate on a wood floor, with a force of 50N. If the crate has a mass of 25kg, and µs = 0.5, will the crate move?

Ff,s = µs × FN.

Static Friction can vary between 0 and this maximum value

Ex3) A person pushes on a wooden crate on a wood floor, with a force of 50N. If the crate has a mass of 25kg, and µs = 0.5, will the crate move? FN

Ff,s = µs × FN. = (.5)(250N)

Ff,s F = 50N = 125N

Friction depends on: 1. the types of surfaces (µ) 2. the force pressing the surfaces together (FN)

(not speed or surface area)

Lab ex) Your group weighs a block with a spring scale and finds it to be 15N. The force required to pull it at constant speed is 12N. What is the coefficient of friction between the block and table?

Ch6 HW #3 14 – 17

Lab ex) Your group weighs a block with a spring scale and finds it to be 15N. The force required to pull it at constant speed is 12N. What is the coefficient of friction between the block and table?

FN Ff,k = µk × FN

12N = (µk)(15N) Ff,s F = 12N

µk = 0.8 Fg

Ch6 HW #3 14 – 17

Lab6.3A – Calculating Coefficients of Friction

- due tomorrow

Ch6 HW#3 14 – 17 14. Calculate the kinetic friction for the rubber soled shoe on wet concrete.

mass of 55kg, µk = 0.40.

15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes if the friction force required to get you sliding is 350N.

FN Ff,k = µk × FN

Ff,k = (.4)(550N) Ff,k F

Ff,k = 220N Fg 15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes

if the friction force required to get you sliding is 350N.

FN Ff,k = µk × FN

Ff,k = (.4)(550N) Ff,k F

Ff,k = 220N Fg 15.What is the coefficient of static friction for you (55kg) in your rubber soled shoes

if the friction force required to get you sliding is 350N.

FN Ff,s = µs × FN

350N = (µs)(550N) Ff,s F = 350N

µs = 0.64 Fg

16. A person pushes on a wooden crate on a wood floor. Crate has a mass of 45kg, µs = 0.50, what force necessary to get it started moving?

17. Your lab group weighs a block with a spring scale and finds it weighs 9.0N.The force required to move it at constant speed across light sand paper is 15.5N. What is µk?

FN Ff,s = µs × FN

Ff,s = (.5)(450N) Ff,s F

Ff,s = 225N Fg

FN Ff,s = µs × FN

Ff,s = (.5)(450N) Ff,s F

Ff,s = 225N Fg

FN Ff,k = µk × FN

15.5N = (µk)(9N) Ff,k F

µk = 1.72 Fg

Friction- Day 2

Ex1) A lab group pulls a 6kg ball at constant speed across a table with a force of 15N.What’s the coefficient of friction?

F= 15NFf,k

Friction- Day 2

Ex1) A lab group pulls a 6kg ball at constant speed across a table with a force of 15N.What’s the coefficient of friction?

F= 15NFF,K

Fnet = F – Ff,k

m.a = 15N – μk ×FN

0 = 15 – μk ×60N

μk = .25

Ex2) Force of 2.5N is required to get a .4kg. block sliding. A force of 2N is required to keep it moving at a constant speed. Find µs and µk.

F = 2.5NFF,S FF,S= μs×FN

2.5N= μs × 4N μs= .63Fg= 4N

Ex2) Force of 2.5N is required to get a .4kg. block sliding. A force of 2N is required to keep it moving at a constant speed. Find µs and µk.

F = 2.5N

Fg= 4N

FF,S FF,S= μs×FN

2.5N= μs × 4N μs= .63

Fg= 4N

F = 2NFF,K FF,K= µk × FN

2N = µk × 4N μ = .5

Ex3) A person pushes a crate of 20kg with a force of 101N. μs = 0.50. Will the Crate move?

Ex3) A person pushes a crate of 20kg with a force of 101N. μs = 0.50. Will the Crate move? FN

Ff,s = μs x FN

Ff,s F=101N Fg=200N = (.5)(200N)

= 100N Yes!

What force will be needed to keep it moving if μk = 0.20?

Ff,s = μs x FN

Ff,s F=101N Fg=200N = (.5)(200N)

= 100N Yes!

Ff,k = μk x FN

= (.2)(200N) = 40N

If the person still pushes with 101N, what is the acceleration of the crate?

Ff,s = μs x FN

Ff,s F=101N Fg=200N = (.5)(200N)

= 100N Yes!

Ff,k = μk x FN

= (.2)(200N) = 40N

If the person still pushes with 101N, what is the acceleration of the crate?

Fnet = F – Ff,k m.a = 101N – 40N

(20kg).a = 61N a = 3.1 m/s

Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet , and rolls 5.5m during a time of 3.0sec. Find the friction between the carpet and ball.

(Takes 4 steps:)

1. d = ½(vi+vf).t

2. vf = vi + at

3. F = m.a

4. Ff,k = μk x FN

a = -___

d= 5.5m vf= 0 t= 3 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 5.5 = ½(vi + 0)3 vi = 3.7 m/s

2. vf = vi + at

3. F = m.a

4. Ff,k = μk x FN

a = -___

d= 5.5m vf= 0 t= 3 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 5.5 = ½(vi + 0)3 vi = 3.7 m/s

2. vf = vi + at 0 m/s = 3.7 m/s + a(3s) a = - 1.23 m/s2

3. F = m.a

4. Ff,k = μk x FN

a = - 1.23 m/s2

d= 5.5m vf= 0 t= 3 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 5.5 = ½(vi + 0)3 vi = 3.7 m/s

3. F = m.a F = (0.25kg)(-1.23 m/s2) F = - 0.3N

4. Ff,k = μk x FN

a = - 1.23 m/s2

d= 5.5m vf= 0 t= 3 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 5.5 = ½(vi + 0)3 vi = 3.7 m/s

3. F = m.a F = (0.25kg)(-1.23 m/s2) F = - 0.3N

4. Ff,k = μk x FN 0.3N = μk x (2.5N) μk = 0.12

a = - 1.23 m/s2

d= 5.5m vf= 0 t= 3 sec.

Ch6 HW#4 18,19

Lab6.3B – Sliding Friction

- due tomorrow

Lab6.3C – Friction on Carpet

- due tomorrow

Ch6 HW#4 18,1918. Ex3) A person pushes a crate of 30kg with a force of 25N. μs = 0.78.

Will the Crate move? FN

Ff,s = μs x FN

Ff,s F=25N Fg=300N

Will the crate move if μs = 0.04?

If the person still pushes with 25N, what is the acceleration of the crate? (μk = 0.04 also)

Ff,s = μs x FN

Ff,s F=25N Fg=300N = (.78)(300N)

= 234N No!

Ff,s = μs x FN

Ff,s F=25N Fg=300N = (.78)(300N)

= 234N No!

Ff,s = μs x FN

= (.04)(300N) = 12N Yes!

Ff,s = μs x FN

Ff,s F=25N Fg=300N = (.78)(300N)

= 234N No!

Ff,s = μs x FN

= (.04)(300N) = 12N Yes!

Fnet = F – Ff,k m.a = 25N – 12N

(30kg).a = 13N a = 0.43 m/s

19. A 14kg ball rolls off a ramp and down a piece of carpet , and rolls 25.2m during a time of 4.5sec. Find the friction between the carpet and ball.

(Takes 4 steps:)

1. d = ½(vi+vf).t

2. vf = vi + at

3. F = m.a

4. Ff,k = μk x FN

a = - ____ m/s2

d= 25.2m vf= 0 t= 4.5 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 25.2 = ½(vi + 0)4.5 vi = 11.2 m/s

2. vf = vi + at

3. F = m.a

4. Ff,k = μk x FN

a = - ____ m/s2

d= 25.2m vf= 0 t= 4.5 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 25.2 = ½(vi + 0)4.5 vi = 11.2 m/s

2. vf = vi + at 0 m/s = 11.2 m/s + a(4.5s) a = - 2.4 m/s2

3. F = m.a

4. Ff,k = μk x FN

a = - ____ m/s2

d= 25.2m vf= 0 t= 4.5 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 25.2 = ½(vi + 0)4.5 vi = 11.2 m/s

3. F = m.a F = (0.25kg)(-2.4 m/s2) F = - 33.6N

4. Ff,k = μk x FN

a = - 2.4 m/s2

d= 25.2m vf= 0 t= 4.5 sec.

(Takes 4 steps:)

1. d = ½(vi+vf).t 25.2 = ½(vi + 0)4.5 vi = 11.2 m/s

3. F = m.a F = (0.25kg)(-2.4 m/s2) F = - 33.6N

4. Ff,k = μk x FN 33.6N = μk x (140N) μk = 0.24

d= 25.2m vf= 0 t= 4.5 sec.

a = - 2.4 m/s2

Ch6.4 – Periodic MotionSimple harmonic motion – An object oscillates back and forth.Period - how long it takes to complete one cycleAmplitude - Its maximum distance from rest position.Equilibrium - rest position, all forces cancelling out.

1. A Pendulum

Period of a pendulum:

FgFNet

1. A Pendulum

Depends on string lengthPeriod of a pendulum:

and accl of gravity.

Pendulums don’t care about the mass attaached!

FgFNet

Ex1) What is the length of a pendulum with a period of 2 seconds?

99.8.9

2. Springs – amount they stretch is proportional to the mass attached.

F = k.d

elastic spring distanceforce constant displaced

Ex2) A 100g mass is attached to a spring and it stretches 10 cm.What is the value of the spring constant?

Period of a Spring

springs care about the mass attached

and the stretchiness of the spring

to determine the time to oscillate.

Ex3) What is the spring constant for a spring that has a period of 0.63s when a mass of 100g is attached?

Period of a Spring

springs care about the mass attached

and the stretchiness of the spring

to determine the time to oscillate.

Ex3) What is the spring constant for a spring that has a period of 0.63s when a mass of 100g is attached?

Resonance – with a small force applied, an object will oscillate at its natural frequency. Keep applying force, amplitude increases tremendously.

Ch6 HW#5 20 – 24

1263.0

Lab6.4 – Periodic Motion

- due tomorrow

Ch6 HW#5 20 – 24 20. Period of a pendulum with length of 0.10m?

21. Length of pendulum with period of 5s?

22. Accl of gravity on planet where pendulum of 1m length has a period of 1 sec?

23. Mass of 0.250kg suspended from a spring that stretches 0.05m. Find spring constant.

24. A 0.01kg mass suspended from a spring with a spring constant of 10 N/m. How much stretch?

Ch6 HW#5 20 – 24 20. Period of a pendulum with length of 0.10m?

21. Length of pendulum with period of 5s?

22. Accl of gravity on planet where pendulum of 1m length has a period of 1 sec?

23. Mass of 0.250kg suspended from a spring F = k.dthat stretches 0.05m. Find spring constant. 2.5N = k.(.05m)

24. A 0.01kg mass suspended from a spring with F = k.da spring constant of 10 N/m. How much stretch? 0.1N = (10N/m).(d)

10.022

Ch6.5 – Newton’s 3rd Law - For every action (force) there is an equal and opposite reaction (force)

- Forces always come in pairs (Interaction pairs)

FA on B = FB on A

Ex1) A 100kg football player pushes off his 70kg girlfriend while ice skating. If her acceleration is 6 m/s

2, what is his?

Ch6.5 – Newton’s 3rd Law - For every action (force) there is an equal and opposite reaction (force)

- Forces always come in pairs (Interaction pairs)

FA on B = FB on A

Ex1) A 100kg football player pushes off his 70kg girlfriend while ice skating. If her acceleration is 6 m/s

2, what is his?

FG on B = FB on G

mB × a= mG × a100a = (40)(6)

a= 2.4

Ex2) A 50kg person jumps upwards off Earth’s surface w/ an acceleration of 3 m/s2.

What is the acceleration of the Earth downward? (ME= 6×1024 kg)

Ex2) A 50kg person jumps upwards off Earth’s surface w/ an acceleration of 3 m/s2.

What is the acceleration of the Earth downward? (ME= 6×1024 kg)

FE on P = FP on E mp×a = mE×a

(50)(3) = (6×1024)(a) a = 2.5×10-23 m/s

Ex3) A 300kg astronaut is skydiving on the moon. If he free falls at 1.6 m/s2, how fast does the moon accelerate up? (Mm = 7×1022 kg)

FM on P = FP on M mp×a = mM×a

(300)(1.6) = (7×1022)(a) a = 7×10-21 m/s

Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s. By how much does the earth accelerate its spin?

FT on E FE on T

vf = vi + at 12m/s = 0 + a(3)

FT on E FE on T a = 4 m/s2

vf = vi + at 12m/s = 0 + a(3)

FT on E FE on T a = 4 m/s2

FE on T = FT on E mT×a = mE×a

(1000)(4) = (6×1024)(a) a = 7×10-22 m/s

Ch6 HW#6 25 – 28

Ch6 HW#6 25 – 28 25. 7kg rock free falls, what is accl of earth?

26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of 20 m/s2,

what is accl of asteroid?

FE on R = FR on E mR×a = mE×a

7)(9.8) = (6×1024)(a) a = 1×10-23 m/s

26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of -20 m/s2,

FE on R = FR on E mR×a = mE×a

7)(9.8) = (6×1024)(a) a = 1×10-23 m/s

26. 8000kg truck brakes on 50,000kg asteroid with a deceleration of -20 m/s2,

FA on T = FT on A mT×a = mA×a

(8000)(-20) = (50000)(a) a = -3.2 m/s

27. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m/s2, what is his accl?

28. What is the force on the boy?

FG on B = FB on G mB×a = mG×a

(65)(a) = (45)(3) a = 2 m/s

vi = 30m/svf = 0m/st = 0.005sa = ?

(65)(a) = (45)(3) a = 2 m/s

vi = 30m/s vf = vi + atvf = 0m/s 0 = 30 + a(0.005)t = 0.005s a = -6000 m/s

(65)(a) = (45)(3) a = 2 m/s

vi = 30m/s vf = vi + atvf = 0m/s 0 = 30 + a(0.005)t = 0.005s a = -6000 m/s

2 a = ?

FGlove on Ball = FBall on Glove mB×a = mG×a (0.145)(-6000) =

F = -870N

Ch6.6 – Coupled MotionEx1) Find the acceleration of this system:

Ex2) Find the acceleration of this system:

Ex3) Find the acceleration of this system:

Air Drag- Friction with the air DOES depend on speed

(Faster you fall, the more air has to get out of your way.)- DOES depend on surface area. (Same reason)

Fnet Accl Velocity

Ch6 HW#7 29 – 32

Lab6.5 – Coupled Motion

- due tomorrow

Ch6 HW#7 29 – 3229) Find the acceleration of this system:

Fnet = Fg3 – FT + FT – Fg2 ..

Fnet = Fg3 – FT + FT – Fg2

mT.a = 30N – 20N

(5kg)a = 10N FT

Fg2 a = 2 m/s2

Fnet = Fg5 – FT + FT – Fg3

mT.a = 50N – 30N

(8kg)a = 20N FT

a = 2.5 m/s2

31) Find the acceleration of this system:

Fnet = Fg3 – FT + FT

Fnet = Fg3 – FT + FT Fg3

mT.a = 30N

(5kg)a = 30N

a = 6 m/s2

Fnet = Fg5 – FT + FT

Fnet = Fg5 – FT + FT Fg5

mT.a = 50N

(8kg)a = 50N

a = 6.25 m/s2

Ch6 Rev1. A rock is dropped from a bridge into a valley. The Earth pulls on the rock

and accl it downward. According to Newton’s 3rd Law, the rock must also be pulling on the Earth, yet the Earth doesn’t seam to accl. Explain.

2. State Newton’s 3 Laws.

3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s.a. Find accl vf = vi + at

b. Find net force F = m.a

4. A bowling ball has a mass of 8.0kg. What is its weight?

3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s.a. Find accl vf = vi + at 26.3m/s = 0m/s + a(0.59s) a = 44.6m/s

b. Find net forceF = m.a = (873kg)(44.6m/s

2) = 38,910N

3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s.a. Find accl vf = vi + at 26.3m/s = 0m/s + a(0.59s) a = 44.6m/s

b. Find net forceF = m.a = (873kg)(44.6m/s

2) = 38,910N

F = m.g = (8.0kg)(9.8m/s2)= 78.4N

Ch6 Rev5. Your new motorcycle weighs 2450N. What is its mass?

6. A pendulum has a length of 0.67m. Find its period.

7. What is the length of a pendulum with a period of 2.5s?

F = m.g 2450N= (m)(9.8m/s

2)m = 250kg

F = m.g 2450N= (m)(9.8m/s

2)m = 250kg

F = m.g 2450N= (m)(9.8m/s

2)m = 250kg

55.18.9

Ch6 Rev8. What is the mass of an object hanging from a spring if the period

of oscillation is 3s, and the spring constant has a value of 50 N/m.

9. If a mass weighing 20N is suspended from a spring causing it to stretch15cm, what is the value of the spring constant?

F = k.d 20N = k.(.15m)

k = 133 N/m

Ch6 Rev10. If you use a horizontal force of 30N to pull a 12.0kg wooden crate across

a floor at constant velocity, what is the coefficient of kinetic frictionbetween the crate and floor?

Ff,k = μk x FN

Ff,k F=12N Fg=120N

11. Calculate the kinetic friction force when a 200kg sled slides acrossa frozen lake, coefficient of kinetic friction = 0.06.

Ff,k = μk x FN FN

30N = μk x 120N Ff,k F=30N Fg=120N μk = 0.25

Ff,k = μk x FN FN

Ff,k F=? Fg=2000N

Ff,k = μk x FN FN

30N = μk x 120N Ff,k F=30N Fg=120N μk = 0.25

Ff,k = μk x FN FN

Ff,k = (0.06)(2000)N Ff,k F=? Fg=2000N Ff,k = 120N

Ch6 Rev12. A person pushes a 25kg box with a force of 26N, coefficient of static

friction is 0.43.a. Will the crate move? Ff,s = μs x FN

Ff,s F=26N Fg=250N

b. If coefficient decreases to 0.10, will it move?

c. If the person still pulls with 26N, and the coefficient of kinetic friction is 0.05, what is the accl?

= (.43)(250N) Ff,s F=26N Fg=250N = 108N no!

b. If coefficient decreases to 0.10, will it move?

= (.43)(250N) Ff,s F=26N Fg=250N = 108N No!

b. If coefficient decreases to 0.10, will it move? Ff,s = μs x FN

= (.10)(250N) = 25N Yes!

= (.43)(250N) Ff,s F=26N Fg=250N = 108N No!

b. If coefficient decreases to 0.10, will it move? Ff,s = μs x FN

= (.10)(250N) = 25N Yes!

Fnet = F – Ff,k m.a = 26N – μk x FN

(25kg).a = 13N – (.05)(250N) a = 0.02m/s

Ch6 Rev13. A 60kg skydiver jumps from an airplane.

a. What is the force of the earth on skydiver?

b. What is the force of the skydiver on earth?

c. If the mass of the earth is 6x1024kg, what is its accl?

a. What is the force of the earth on skydiver?FE on P Fg = m.g

= (60kg)(9.8 m/s2)

= 588N FE on P

b. What is the force of the skydiver on earth? FP on E

= (60kg)(9.8 m/s2)

= 588N FE on P

b. What is the force of the skydiver on earth? FP on E FP on E = FE on P

FP on E = 588N (Same!)

= (60kg)(9.8 m/s2)

= 588N FE on P

b. What is the force of the skydiver on earth? FP on E FP on E = FE on P

FP on E = 588N (Same!)

FP on E = mE.a

588N = (6x1024kg).a a = 9.8x10-23 m/s

Ch6 Rev14. What is the accl of this system? 15. What is the accl of this system?

Fnet = Fg3 – FT + FT – Fg1 Fnet = Fg7 – FT + FT – Fg2

mT.a = 30N – 10N

(4kg)a = 20N FT

Fg1 a = 5 m/s2

Fnet = Fg3 – FT + FT – Fg1 Fnet = Fg7 – FT + FT – Fg2

mT.a = 30N – 10N mT

.a = 70N – 20N FT FT

(4kg)a = 20N (9kg)a = 50N FT FT

Fg1 a = 5 m/s2 Fg2 a = 5.6 m/s

Fg3 Fg7

Lab6.5A – Newton’s Third Law – Balloon Rockets

Materials: 1 balloon, one straw cut to any length, one piece of tape, and a long string.Procedure: 1. Your lab group decides on the design. Use the low friction string first. 2. Once you are successful in getting the balloon across the room on the string, repeat the procedure on the thick string.Analysis: 1. Use Newton’s third law to explain how the balloon rocket is able to move

across the room. Include a free-body diagram that shows the interaction-pair of forces exhibiting Newton’s third law.

100 90 80 70vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10

time (sec)

Ffk Fhangar

Ch6.1 - Forces Newton's 1st Law - object in motion will remain in that motion, an object at...

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