Work & EnergyDay #1: Introduction

I. A. Define Work:

Work is a measure of what is accomplished when a force is applied onto an object while the object moves. The object must be moving while the force is applied.

B. Equation of work:

Work is defined as the product of the force applied on an object and the displacement of the object.

cosxFW

cosxFW

F = force acting on the object

x = displacement of the object

The angle is the angle between the direction of the force F and the displacement x.

C. Units for work:

The product of force and displacement gives units of newtons times meters.

This unit is given a special name: joule = J

The relationship is given as follows:

mNJ 11

ms

mkgJ

2112

2

11s

mkgJ

D. Alternate expression for work:

cosxFW

The term F||=Fcos is the component of the force F that is parallel to the displacement x. The work is the product of these two parallel components.

xFW ||

E. Work and sign (±):

cosxFW xFW ||

For the equation of work on the left, there is a dependence on the direction of the force relative to the displacement. If 0o ≤ < 90o, then the work is positive. If 90o < ≤ 180o, then the work is negative.

If F|| points in the same direction as the displacement x, then the work is positive.If F|| points in the opposite direction as the displacement x, then the work is negative.

Ex. #1: A 10.0 kg mass is pulled towards the right 10.0 m by an applied force of 40.0 N. The applied force also points towards the right and the object moves at a constant velocity. (a) Draw all the forces acting on the object.

n = Normal Force

mg = Weight

Fo = Applied Force

Fk = Kinetic Friction Force

x = Motion

cosxnWnormal

(b) What is the work done by the normal force?

oxn 90cos 0

(c) What is the work done by the weight force of the object?

cosxmgWweight oxmg 90cos 0

General Rule: Any force applied perpendicular to the motion does not do any work!

(d) What is the work done by the applied force?

cosxFW oapplied

0cos0.100.40 mNWapplied

JWapplied 400

(e) What is the work done by the friction force?

Fo = Applied Force

Fk = Kinetic Friction Force

Since the object moves at a constant velocity, the forces balance! NFF ko 0.40

cosxFW kfriction

ofriction mNW 180cos0.100.40 J400

(f) What is the net work on the object?

frictionappliedweightnormalnet WWWWW

JJWnet 40040000

0netW

0netF

Ex. #2 Is it possible to do work on an object that remains at rest? Why of why not?

NO. Work requires a displacement: cosxFW

Ex. #3: A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

1) friction does no work at all

2) friction does negative work

3) friction does positive work

cosxFW

Friction points opposite to the motion of the object!

0180cos oxFW

Ex. #4: Can friction ever do positive work?

YES! Consider a box placed on the back of a flatbed truck:

When the truck drives forwards, the inertia of the box makes the box appear to slide to the back of the truck.

Friction between the box and the truck pulls the box forward, opposing its motion towards the back of the truck.

Fs

Ex. #5: In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball?

1) catcher has done positive work on the ball

2) catcher has done negative work on the ball

3) catcher has done zero work on the ball

The force exerted by the catcher is oppositeopposite in in direction to the displacement of the ball, so the direction to the displacement of the ball, so the work is negativework is negative. Or using the definition of work (WW = = F d cos F d cos ), since = = 180180oo, then W < W < 00. Note that because the work done on the ball is negative, its speed decreases.Follow-up:Follow-up: What about the work done by the ball on the catcher? What about the work done by the ball on the catcher?

Ex. #6: A man with a weight of 735 N stands in an elevator that accelerates upwards at 1.20 m/s2 for 5.00 m. (a) What is the work done by gravity?

n = Norm

al Force

mg = Weight

cosxmgWweight

NOTE: Since the object moves upwards, and the weight points downwards, = 180o.

oweight mNW 180cos00.5735

3680weightW J

(b) What is the work done by the normal force?

NOTE: Since the acceleration is given, solve for the normal force using Newton’s 2nd Law:

maFnet

The sum of the forces is: mgnFnet

mamgn mgman

NOTE: the weight is given, not the mass! Nmg 735

kgNms

m0.75

80.9735

2

mgman

Nkgn sm 73520.10.75 2

Nn 825

cosxnWnormal NOTE: Since the object moves upwards, and the normal force points upwards, = 0o.

0cos00.5825 mNWnormal

4130normalW J

Ex. #7: A block is pulled towards the right with a force of 50.0 N applied at an angle of 36.9o above the horizontal. The block is already in motion and moves a distance of 20.0 meters at a constant velocity.(a) What is the work done by the applied force?

m

mg = weight

n = normal Fo

= 36.9oFK

cosxFW oapplied

oapplied mNW 9.36cos0.200.50 J800

(b) What is the work done by the force of friction?

The object moves at a constant velocity, which means the forces balance. Friction will balance the horizontal component of the applied force.

ook NFF 9.36cos0.50cos N0.40

ok mNW 180cos0.200.40

mNWk 0.200.40 J800

Note that the net work is zero when the velocity is constant.

(c) If the block has a mass of 12.0 kg, what is the coefficient of kinetic friction?

m

mg = weight

n = normal Fo

= 36.9oFK

mgFnF oynet sin0

sinoFmgn

nF kk

sino

kkk Fmg

FnF

457.0

Ex. #8: A mass of 44.0 kg is pulled towards the top of a ramp at a constant speed by an applied force. The applied force is parallel to the ramp, and points towards the top of the ramp. The coefficient of friction between the ramp and the surface is 0.400 and the block is pulled a distance of 5.00 meters up the ramp. What is the work done by each force? What is the total work on the object? The angle of the incline is 30.0o.

m

mg

cosmg

sinmg

oFn

kF

The first step is to solve for each force:

The ┴ forces balance: cosmgn

From the definition of kinetic friction:

nF kk

definition

cosmgk

substi

tute

The net force parallel to the ramp is zero as well:

||0 netF sinmgFF ko

subs

titut

e sincos mgmgF ko

The work done by the normal force is zero:

cosxnWnormal oxn 90cos 0

Calculate the work done by gravity through its components. Only the parallel component will do work.

ogravity xmgW 180cossin

mkgW os

mgravity 00.50.30sin80.90.44 2

JWgravity 1078

The work done by the applied force is:

ooo xFW 0cos

sincos mgmgF ko

oos

mo kgF 0.30sin0.30cos400.080.90.44 2

365oF N

365 5.00 cos 0ooW N m J1825

The work done by the friction force is:

okdrag xmgW 180coscos

mkgW os

mdrag 00.50.30cos80.90.44400.0 2

JWdrag 747

Finally, the total work done is:

JJJWnet 18257471078 0netW

Work & EnergyDay #2: Work Energy Theorem

A 10.0- kg block of wood rests on the lab bench. A force of 50.00 N force is applied at a 30.0 degree angle above the horizontal and it moves with constant velocity.

1.Calculate the weight.2.Calculate the normal force.3. How much work is done to move the block 1.00 meters?4. Calculate the frictional work.5. Calculate the coefficient of friction.

1. W = mg = 98N

2. N= mg- Fapplied X Sinɵ = 73 N

5. µ = Ff/N = = 43N/73N= 0.59

3. Wapplied = F‖X distance = FappliedXCosɵ X 1 m = 43 J4. Wfriction = - W applied = -43 J

Work Energy TheoremI. Introduction:A. What is energy?

Energy is defined as the ability to do work.

If an object has energy, then this object can perform work.

Energy and work are two forms of the same overall concept.

B. What are the types of energy?1. kinetic energy:

Kinetic Energy is defined as energy associated with motion. Work can be done from the movement of an object. wrecking

ballwater wheel

2. potential energy:

Potential Energy is defined as energy associated with position. Energy can be stored in an object by virtue of its position.

pile driver pile driver #2

C. Equation and units for kinetic energy:

Definition of kinetic energy KE:2

21 mvKE

follow the equation for units:

221 speedmassKE

2

smkg

2

2

smkg J

work and energy are two parts of the same overall concept, therefore they share the same units

D. What is the work energy theorem? WWKE net

The work energy theorem states that the change of the kinetic energy of an object is equal to the total amount of work done on the object.

Note: the net work can be found by either finding the sum of the work done by each individual force or by finding the work done by the net force.

321cos WWWWxFW netnet

II. Examples:Example #1: What is the kinetic energy of a 10.0 kg mass moving at 6.00 m/s?

221 mvKE 2

21 00.60.10 s

mkg J180

Ex. #2: What speed should a 2.50 kg mass have so that it has the same kinetic energy as the above example?

mKEv 2

kgJ

50.21802

sm0.12

fast solution: m and v2 are inversely proportional.

4m implies 42v so finally 2v

Ex. #3: {on your own} If a third mass has a speed of 4.00 m/s, what should its mass be so that it has the same kinetic energy as that of problem #1?

221 mvKE

2

2vKEm

kgJmsm

5.2200.41802

2

Ex. #4: A force of 20.0 N pushes a 10.0 kg mass for a distance of 10.0 meters. The surface is frictionless. What is the speed of the object if it starts from rest?

cosxFW oo

oo mNW 0cos0.100.20 J200

KEWnet oKEKE 2212

21

omvmv

0

mWv net2

kgJ

0.102002

sm32.6

Ex. #5: {on your own} How much force is needed to push a 4.00 kg mass from a speed of 5.00 m/s to a speed of 7.00 m/s in a distance of 5.00 meters?

KEWnet oKEKE 2212

21

omvmv

2221 00.500.700.4 s

msm

net kgW

JWnet 0.48 oo xF 0cos

NmJFo 60.9

00.50.48

Ex. #6: A mass of 10.0 kg is raised by a rope with a force of 110 N.a. What is the work done by this force if the mass is raised upwards 2.40 meters?

cosxFW oo omN 0cos40.2110

b. What is the work done by gravity?

cosxmgWgravity

os

mgravity mkgW 180cos40.280.90.10 2

235gravityW J

= 264 J

c. What is the net work on the object?

gravityonet WWW JJ 2.235264

JWnet 8.28 Wnet > 0, object speeds up

d. What is the speed of the object at the end of the motion if the mass starts from rest?

KEWnet oKEKE 2212

21

omvmv 0

mWv net2

kgJ

0.108.282

sm40.2

Ex. #7: A 4.00 kg mass has an initial upward velocity of 14.0 m/s. A cord is lifting upwards on the mass. If the mass slows uniformly to a stop in a distance of 40.0 meters, what is the lifting force?

KEWnet oKEKE 2212

21

omvmv 0

JWnet 392

gravityTnet WWW

oonet xmgxTW 180cos0cos

xmgTWnet

oonet xmgxTW 180cos0cos

mgx

WT net

280.900.40.40

392s

mkgmJ

NT 4.29

Ex. 8: A 3000 kg car has an applied force of 2500 N moving the car forwards. At the same time, there is a 1500 N resistive force acting on the car. a. How much distance is needed to accelerate the car from 20.0 m/s to 25.0 m/s?

NFo 2500NFdrag 1500

dragonet FFF NN 15002500

NFnet 1000

KE 2212

21

omvmv

2221 0.200.253000 s

msmkg

kJJKE 5.337500,337

onetnet xFWKE 0cos

mN

JFKExnet

3381000

500,337

b. How much braking force is needed to slow the car from 25.0 m/s to a stop in a distance of 100 meters? What is the needed coefficient of static friction?

mg

n

NFdrag 1500

?brakeF

KE 2221 0.2503000 s

msmkg

JKE 500,937

mg

n

NFdrag 1500

?brakeF

mgn mgnF sss

obrake

odragnet xFxFWKE 180cos180cos

xFxFKE brakedrag

brakedrag FFx

dragbrake Fx

KEF

Nm

JFbrake 1500100

500,937

NFbrake 7875

mgFF ssbrake

mgFbrake

s 280.930007875

smkg

N

268.0s

Ex. #9: If the speed of an object is doubled from speed v to speed 2v, what happens to the kinetic energy of the object? How does the distance compare for pushing the object from rest to speed v as compared to pushing the object from rest to speed 2v?

2221 vmvKE

2v implies 42v so that 4KE

For accelerating an object:

cos2212

21 xFWKEmvmv oneto

For accelerating from 0 → v:

o

o

Fvvmx

2

22

0

oFmv2

2

For accelerating from 0 → v compared to 0 → 2v:

2v implies 42v so that 4x

Ex. #10: A 1.80-kg particle has a speed of 2.0 m/s at point A and a kinetic energy of 22.5 J at point B. What is (a) its kinetic energy at A? (b) its speed at point B? (c) the total work done on the particle as it moves from A to B?

JkgmvKE sm

A 60.30.280.1 2212

21

smB

B kgJ

mKEv 00.5

80.15.2222

JJKEKEW ABBA 60.35.22

JW BA 9.18

Work & EnergyDay #3: Conservation of Energy

HW:Section #3

Potential Energy and Conservation of Energy.I. Introduction.A. What is potential energy?

Potential Energy is defined as energy associated with position. Energy can be stored in an object by virtue of its position.

B. What is the difference between a conservative and non–conservative force?

Work is calculated by applying a force on an object from some starting point to some finishing point.

It is possible to take different paths from a given starting point to a given ending point.

The work done by a conservative force is independent of the path taken from start to finish.

A non – conservative force is not conservative, the work does depend on the path taken.

C. How is potential energy defined?

Potential energy is defined for conservative forces only.

cWPEU

PE = change of potential energy between two points. {some books use U instead of PE}

Wc = work done by some conservative force between the same two points {start to finish}

D. Derive the potential energy of gravity near Earth’s surface.

Work done in lifting an object upwards:

180cosxmgWgravity

mghWgravity

PE equals the minus of the work:

gravityWPE mgh mgh

Potential energy increases as an object rises upwards.

Work done in lowering an object downwards:

0cosxmgWgravity

mghWgravity

PE equals the minus of the work:

gravityWPE mgh mgh

Potential energy decreases as an object lowers downwards.

E. Derive the work done by gravity and the potential energy of gravity on an incline.

mg

cosmgF

sin|| mgF

The motion of the mass will be parallel to the ramp.

Only the parallel component of gravity will do work.

xh

xh

sin

For motion to the top of the ramp:

xFWgravity ||

If F|| points in the opposite direction as the displacement x, then the work is negative.

xmgWgravity sin xx

hmg

mghWgravity This is the same result as lifting the object straight upwards!

gravityWPE mgh mgh

The change in the PE only depends on the vertical height gain, not the path! conservative

In General: Potential energy increases with height.

Define the PE for any height as follows:

Let the y – axis be the vertical axis, and pick some point to be the origin. This is the point of zero potential energy.

The PE at any height y is given as: mgyPE

Since y can be negative, PE can be negative.

Example: The student has hit rock bottom (y = 0) and is showing signs of digging.

II. Examples.Ex. #1: A ball of mass 0.500 kg has a potential energy of 49.0 J relative to the ground.a. How high above the ground is the ball initially?

Let the ground level be zero height.

mgyPE

mgPEy 280.9500.0

0.49

smkg

J

my 0.10

b. How much kinetic energy will the ball have when it reaches the bottom? How fast will the ball be traveling?

The ball loses 49.0 J of PE as it falls. JPE 0.49

JWPE gravity 0.49

JWgravity 0.49

ogravity KEKEJWKE 0.490

JKE 0.49 and

kgJ

mKEv

500.00.4922

smv 0.14

Ex. #2: A mass of 4.00 kg slides down an incline that measures 2.40 meters tall and 4.00 meters along the ramp. The mass starts from rest and the ramp is frictionless. a. What is the starting potential energy of the mass, relative to the ground?

PE only depends on vertical height. y = 0 at bottom.

mkgmgyPE sm 40.280.900.4 2

94.1PE J

b. What is the kinetic energy and speed of the mass at the bottom of the ramp?

The ball loses 94.08 J of PE as it falls. 94.1PE J

94.1gravityPE W J

94.1gravityW J

94.1gravity oKE W J KE KE 0

94.1KE J and 2 94.124.00

JKEvm kg

smv 86.6

III. Force and potential energy stored in a spring.

A spring is known as a linear restoring force.

xi is the location where the end of the spring would sit if there were no forces applied to pull or push the spring.

This is the equilibrium point.

A linear force means the strength of the spring’s push or pull is proportional to the distance x the end of the spring is pulled from equilibrium.

The equation of the force is written in the form:

kxF

F = force from springx = distance from equilibriumk = spring constant

The spring constant is the relative stiffness of the spring, measured in newtons per meter, N/m.

A restoring force always pulls (or pushes) the end of the spring back to the equilibrium point, to restore the original position of the end of the spring.

Stretching or compressing a spring stores energy into the spring.

The amount of energy stored in a spring is:

221 kxPE

Omit next several slides.

B. Include nonconservative forces in the statement of conservation of energy.

ncif WEE

Note: For solving problems, anytime an “applied force” is given, the work from this force can by lumped into the work done by nonconservative forces.

V. Examples:Ex. #3: A spring requires a force of 40.0 N to compress it 5.00 cm. A 200 gram mass is placed against the compressed spring and released from rest. All surfaces are frictionless.a. What is the spring constant of the spring? kxF

xFk

mN

0500.00.40

mN800

b. What is the potential energy stored in the spring?

221 kxPE 2

21 0500.0800 m

mN

JmNPE 00.100.1

c. What is the speed of the projectile when it is released by the spring?

iii PEKEE 0

fff PEKEE 0

221 mvKEPE fi

mPEv i2

kgJ

200.000.12

sm16.3

Tuesday- Do the next problem as a warm up, after students have prepared HW#3 as Table Groups and have presented it.

Ex. #4: A 20.0 gram projectile is compressed against a spring with a spring constant of 400 N/m. The spring is compressed 15.0 cm and the projectile is launched vertically upwards. How high above its starting position will the mass rise?

Warm Up!Hint: Spring P.E becomes Gravitational P.E.

ncif WEE

if PEPE

221

if kxmgy

mgkxy i

f 2

2

my f 0.23

Discuss Pendulum: No Derivation

Ex. #5: A pendulum is made of a heavy mass attached to and suspended from a long, light cord. The cord for this pendulum measures 2.00 meters long and the mass, or “bob”, measures 4.00 kg. Take the lowest point for the mass as zero potential energy. The pendulum is pulled to an initial angle of 30.0o from vertical and released from rest.a. Write an equation that gives the height of the pendulum mass from the lowest point in terms of the length of the cord and the initial angle from the vertical.

cosLhL

cosLLh

cos1Lh

b. What is the initial potential energy of the pendulum mass?

cos1 mgLmghPEi

os

mi mkgPE 0.30cos100.280.900.4 2

JPEi 5.10

c. What is the speed of the mass at the bottom of the swing of the pendulum?

starting PE turns into KE at the bottom of the swing.

bottomtop KEJPE 5.10

221 mvKEbottom

mKEv bottom2

kg

Jv00.4

5.102 s

m29.2

d. How far will the pendulum swing to the other side after passing the lowest point?

Since no energy is lost, the starting PE and the ending PE at the extremes of the swing are equal.

Thus the heights are equal and the angles are equal.

Ex. #6: An Atwood’s machine has a mass, m1, of 10.0 kg suspended on the left and a mass, m2, of 12.0 kg suspended on the right. If the heavier mass begins 2.00 m above the ground and the lighter mass starts at ground level, how fast will the system be traveling when the heavier mass reaches the ground?

m1

m2

0bottomy

hytop

Initially everything starts from rest, so no initial KE. Only m2 has PE, as it is above the ground.

At the end of the motion, both masses are moving, so both objects have KE. Only m1 is above the ground, so only it has PE.

ghmvmvmEghmE endstart 12

2212

121

2

222

1212

112 vmvmghmghm

12

122 2mm

ghmmv

smv 89.1

VI. Examples with Nonconservative Forces.Ex. #7: A 2.00 kg mass is attached to a spring with a spring constant of 500 N/m. The spring is compressed 10.0 cm and the mass is released. After passing the equilibrium position, the spring stretches to 4.00 cm when the mass again stops. Determine the friction force and the coefficient of kinetic friction for the mass on the support surface.

motion

10 cm compression

equilibrium

4 cm extension

displacement = 14 cm

Fdrag

odragnc xFW 180cos

Start with energy conservation: ncif WEE

The mass starts and finishes at rest, so the initial and final energy only consist of potential energy.

ncif WPEPE

xFxkxk dragif 2212

21

x

xxkF fi

drag

2

22

NFdrag 0.15

Fdrag

n

mg

vertical forces balance

mgn

by definition: nF kdrag mgk

mgFdrag

k 765.0

Ex. #8: A mass of 10.0 kg starts from rest at the top of a frictionless ramp 5.00 meters tall. At the bottom of the ramp, the mass slides on a horizontal surface that has a coefficient of kinetic friction of 0.300.a. What is the speed of the mass at the bottom of the ramp?

Initial energy = PE

Final energy = KE

ncif WEE 0

mghPEKEmv if 221

ghv 2 sm90.9

Fdrag

n

mg

vertical forces balance

mgn

by definition: nF kdrag mgk

motion

xmgxFW ko

dragnc 180cos

b. How far along the surface will the mass travel before it comes to a stop?

ncif WEE

ncbottomstop WEE

no height change along flat ground, no change to PE

xmgKEKE kbottomstop 0

xmgmv k 2210

gvx

k2

2

m7.16

Ex. #9: A mass, m1, of 10.0 kg is set on a horizontal table with a coefficient of kinetic friction of 0.200. A light cord ties this mass to another mass, m2, of 6.00 kg suspended over the edge of the table. If the suspended mass falls a distance of 1.00 meter, how fast will the pair of masses be traveling?

m1

m2

m1 does not change height, ignore its PE contribution.

m2 has initial height & PE, but no height and PE at end.

Both masses start from rest and have the same speed at the end. Use energy conservation on the whole system.

( )f i ncE E W m1 slides same distance m2 falls

21

122 2mm

ghmmv k

smv 21.2

PowerI. Introduction.A. Define average power and instantaneous power.B. Equations.

Average Power is defined equal to the work done on an object divided by the elapsed time.

tWPave

Instantaneous Power is rate work is done at one instant of time.

vFP

W = work, t = elapsed time, F = force, v = velocity

C. Units.

tWPave 3

2

2

2 1smkg

ssmkg

sJunits

vFP 3

2

2 smkg

sm

smkg

smNunits

This unit is given its own special name: watt = W

wattWsJ

smNunits

II. Examples.Ex. #1: An automobile engine is capable of producing a maximum of 240 hp. If the car can travel at 115 mph at this full power, how much force is the engine applying towards forward motion of the car?

Horsepower is an older unit of power, and it is equal to 746 W. Convert values to metric:

hp240

hp

W1

746179,000W

hourmile115 51.4 m

s

Use instantaneous power:

vFP vPF

179,000 3480 78351.4 m

s

WF N lbs

Ex. #2: An electric motor on an elevator is rated at 5.00 kW and is used to lift 12,500 N elevator to a height of 100 meters. How much time will it take for the elevator to rise to this height?

Pave = 5.00 kW = 5000 W

F = 12,500 N

x = 100 m

tWPave

txF 0cos

aveP

xFt

W

mNt5000

100500,12 s250

Ex. #3: Bubba notices that his 3000 kg Lincoln slows from 80.0 mph to 60.0 mph in a time of 8.00 seconds if he lets the car coast in neutral. What is the power needed from the motor to keep the car running at a constant velocity of 80.0 mph?

hourmile0.80 35.8 m

s

hourmile0.60 26.8 m

s

Use kinematics to find the acceleration of the car:

atvv o tvva o

26.8 35.88.00

m ms sa

s

21.12 ms

The slowing comes from a friction force:

dragnet FmaF

23000 1.12 mdrag sF ma kg

3350dragF N

The car must produce this same force forwards to maintain a constant 80.0 mph.

The power output is then:

vFP 3350 35.8 msN

120,000P W

Ex. #4: The electric motor of a golf cart can accelerate the golf cart from rest to 10.0 m/s in a time of 5.00 seconds. The mass of the cart with passengers is 700 kg. What is the average power delivered by the motor during the acceleration? What is the instantaneous power at the top speed? If the motor is rated at 25 hp, what is the efficiency of the electric motor at top speed?

Use kinematics to find the acceleration of the car and the distance traveled:

atvv o tvva o

sa s

msm

00.500.10

200.2 sm

2212

21 00.500.20 2 sattvx s

mo m0.25

Average Power:

tWPave

txF 0cos

t

xam

7000aveP W

Instantaneous Power at Top Speed:

vFP

sm

smkgP 0.1000.2700 2

vam

14,000P W

Efficiency:

in

out

PPe

hp

W25000,14

W

hp7461

%1.75751.0 e

Ex. #5: A fully loaded, slow moving freight elevator has a cab with a total mass of 1,200 kg, which is required to travel upward 54 m in 3.0 minutes. The elevator’s counterweight has a mass of only 950 kg. So, the elevator motor must help pull the cab upward. What average power (in horsepower) is required of the force that the motor exerts on the cab via the cable?

The counterweight drops downwards as the elevator rises upwards, so the counterweight lifts on the elevator with a force equal to its weight.

gmgmF cabcwlift

gmgmF cwcablift 280.99501200 smkgkg

NFlift 450,2

The work done is:

mNxFW lift 54450,2 J300,132

The average power is:

min0.3300,132 J

tWP

s60

min1 W735

Converting units:

WhpWP

7461735 hp985.0

Ex. #6: Mike applied 10 N of force over 3 m in 10 seconds. Joe applied the same force over the same distance in 1 minute. Who did more work?

Both exerted the same forcesame force over the

same displacementsame displacement. Therefore, both did the same amount of worksame amount of work. Time Time does not matter for determining the does not matter for determining the

work donework done.

Ex. #7: Mike performed 5 J of work in 10 secs. Joe did 3 J of work in 5 secs. Who produced the greater power?

Since power = work / time, we see that Mike Mike

produced 0.5 Wproduced 0.5 W and Joe produced 0.6 WJoe produced 0.6 W of power. Thus, even though Mike did more work, he required twice the time to do the work, and therefore his power output was lower.

Ex. #8: Engine #1 produces twice the power of engine #2. Can we conclude that engine #1 does twice as much work as engine #2?

No!! We cannot conclude anything about No!! We cannot conclude anything about

how much work each engine does.how much work each engine does. Given the power output, the work will depend upon work will depend upon how much time is usedhow much time is used. For example, engine #1 may do the same amount of work as engine #2, but in half the time.