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Chapter 7Kinetic Energy and Work

Kinetic Energy of a mass m is :

Energy is a scalar quantity The SI unit of energy is

This unit is named after an English scientist of the 1800s, James Prescott Joule.

221mvK

22 s/mkg1J1joule1

2

Sample Problem 7-1In 1896 in Waco, Texas, William Crush of the “Katy” railroad parked two locomotives at opposite ends of a 6.4-km-long track, fired them up, tied their throttles open, and then allowed them to crash head-on at full speed in front of 30,000 spectators.

Hundreds of people were hurt by flying debris; several were killed. Assuming each locomotive weighed 1.2 x 106 N and its acceleration along the track was a constant 0.26 m/s2, what was the total kinetic energy of the two locomotives just before the collision?

3

)xx(a2vv 020

2

)m10x2.3()s/m26.0(20v 322 s/m8.40v

kg10x22.1s/m8.9

N10x2.1m 5

2

6

J10x0.2 8

SOLUTION: 

25221 )s/m8.40()kg10x22.1()mv(2K

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Work

Work W is positive if kinetic energy is transferred to an object by an external force.

Work is negative if kinetic energy is transferred from an object. In other words, negative work is done if an object’s kinetic energy decreases.

Work is a scalar and the SI unit is Joule.

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Work done by an External Force

To calculate the work done on an object by a force during a displacement, we use only the force component along the object's displacement. The force component perpendicular to the displacement does zero work.

cosdFW

dFW

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A force does positive work when it has a vector component in the same direction as the displacement, and it does negative work when it has a vector component in the opposite direction. It does zero work when it has no such vector component.

lbft738.0mN1s/mkgJ1 22

7

WKKK if

particlethe

ondoneworknet

particleaofenergy

kinetictheinchange

donework

netthe

worknetthebefore

energykinetic

doneisworknetthe

afterenergykinetic

WKK if

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Sample Problem 7-2Figure 7-4a shows two industrial spies sliding an initially stationary 225 kg floor safe a displacement of magnitude 8.50 m, straight toward their truck.

The push of Spy 001 is 12.0 N, directed at an angle of 30° downward from the horizontal; the pull of Spy 002 is 10.0 N, directed at 40° above the horizontal. The magnitudes and directions of these forces do not change as the safe moves, and the floor and safe make frictionless contact.

1F

2F

d

9

(a)  What is the net work done on the safe by forces and during the displacement ?1F

2F

d

)30)(cosm50.8)(N0.12(cosdFW 111

J33.88

)40)(cosm50.8)(N0.10(cosdFW 222

J11.65

J11.65J33.88WWW 21

J153J4.153

Work done by :1FSOLUTION: 

Work done by :2F

Total work done :

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(b)  During the displacement, what is the work Wg done on

the safe by the gravitational force and what is the work WN done on the safe by the normal force from the floor?

gF

N

SOLUTION: 

090cosmgdWg

090cosNdWN

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(c)  The safe is initially stationary. What is its speed vf at

the end of the 8.50 m displacement?

SOLUTION: 

kg225

)J4.153(2

m

W2v f

s/m17.1

Work done on object equals increase in kinetic energy :

We have assumed no frictional forces exist.

2i2

12f2

1if mvmvKKW

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Sample Problem 7-3

During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement while a steady wind pushes against the crate with a force

. The situation and coordinate axes are shown in Fig. 7-5.

d

j)N0.6(i)N20(F

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(a)  How much work does this force from the wind do on the crate during the displacement?

i)m0.3(j)N0.6(i)N0.2(dFW

J0.60)1()J0.6(

ij)m0.3()N0.6(ii)m0.3()N0.2(

Work done by the wind force on crate :

SOLUTION: 

The wind force does negative work, i.e. kinetic energy is taken out of the crate.

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(b)  If the crate has a kinetic energy of 10 J at the beginning of displacement , what is its kinetic energy at the end of ?

d

d

J0.4)J0.6(J10WKK if

SOLUTION: 

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Work done by Gravitational Force

The angle = is between (down)

and . For the upward path, .

The work done by the gravitational force is negative for the upward path.

The change of gravitational potential energy of the object equals the negative of the work done by the gravitational force.

cosdgmWg

dgm)1(dgm180cosdgm)pathup(Wg

gF

d

180

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After the object has reached its maximum height and is falling back down, the angle = 0o. The work done is

dgm)1(dgm0cosdgm)pathdown(Wg

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Work done on Lifting and Lowering an Object against a Field at Constant Velocity

An applied force of magnitude = mg pointed up can barely lift an object of weight mg. Assuming no change in the velocity, the work done by the applied force on the object in lifting it a distance d is :

The work done by the applied force in lifting the object is positive. It has the opposite sign as the work done by the gravitational force.

)pathup(WmgdW ga

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The applied force of magnitude mg (pointed up) lowers the object with no change in velocity. The work done by the applied force on the object in lowering it a distance d is :

The work done by the applied force in lowering the object is negative.

The work done by the applied force equals the change in the gravitational potential energy of the object.

mgdWa

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Sample Problem 7-4Let us return to the lifting feats of Andrey Chemerkin and Paul Anderson.

(a)  Chemerkin made his record-breaking lift with rigidly connected objects (a barbell and disk weights) having a total mass m = 260.0 kg; he lifted them a distance of 2.0 m. During the lift, how much work was done on the objects by the gravitational force acting on them?gF

SOLUTION: 

J5100

)180(cos)m0.2()N2548(cosmgd)pathup(Wg

The gravitational force points down and the displacement d points up :

The work done by the gravitational force is negative to the work done by the applied force.

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(b)  How much work was done on the objects by Chemerkin's force during the lift?

SOLUTION: 

J5100)up(W)up(W gAC

(c)  While Chemerkin held the objects stationary above his head, how much work was done on them by his force?

The work done by the applied force is :

Since the displacement d is zero, the work done is zero.

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(d)  How much work was done by the force Paul Anderson applied to lift objects with a total weight of 27 900 N, a distance of 1.0 cm?

SOLUTION: 

J280)m010.0()N27900(

)pathup(W)pathup(W gPA

The work done by the applied force of Paul Anderson is :

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Sample Problem 7-5

An initially stationary 15.0 kg crate of cheese wheels is pulled, via a cable, a distance L = 5.70 m up a frictionless ramp, to a height h of 2.50 m, where it stops (Fig. 7-8a).

(a)  How much work Wg is

done on the crate by the gravitational force during the lift?

gF

23

dsinmg)pathup(Wg

J368

)m50.2()s/m8.9()kg0.15(

mgh)pathup(W2

g

SOLUTION: 

The work done by the gravitational force during the lift is negative :

Since d sin = h, we have :

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(b)  How much work WT is done on the crate by

the force from the cable during the lift?T

SOLUTION: 

Since the crate has zero velocity before and after the lift, the work done by the applied force must be equal and opposite to the work done by the gravitational force.

J368)pathup(WW gT

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Sample Problem 7-6

An elevator cab of mass m = 500 kg is descending with speed vi = 4.0 m/s when its supporting cable begins to slip, allowing it to fall with constant acceleration = /5.

(a)  During the fall through a distance d = 12 m, what is the work Wg done on the cab by the gravitational force ?

a

g

gF

26

kJ59J10x88.5

)1()m12()s/m8.9()kg500(0cosmgd)pathdown(W4

2g

SOLUTION: 

During the fall, the work done on the cab by the gravitational force is positive.

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(b)  During the 12 m fall, what is the work WT done

on the cab by the upward pull of the elevator cable?

T

SOLUTION: 

(positive direction is down)

This gives the magnitude of T, the direction of T is up

The work done on the cab by T during the fall is negative.

dmg)pathdown(W 54

T

kJ47J10x70.4 4

maTmg

mg)ag(mT 54

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(c)  What is the net work W done on the cab during the fall?

SOLUTION: 

kJ12J10x18.1

J10x70.4J10x88.5WWW4

44Tg

(d)  What is the cab's kinetic energy at the end of the 12 m fall?

SOLUTION: 

kJ16J10x58.1

J10x18.1)s/m0.4()kg500(

WmvWKK

4

4221

2i2

1if

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The Spring Force

When a spring of length L0 is either compressed or extended, a restoring force acts opposite to .

This restoring force is known as the spring force and is given by :

The constant k is a scalar and is called the spring constant or force constant. The SI unit is N/m.

The negative sign indicates that the spring force is always opposite in direction to the displacement.

Hooke’s law is named after Robert Hooke of the late 1600s.

d

dkFS

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Work done by the Spring Force

The work done by the spring force as the block is moved from position xi to xf is :

We have assumed that all the mass is at the block and the spring itself is massless.

f

i

x

xS dxFW

f

i

f

i

x

x

x

xS dxxkdx)kx(W

)xx()k(]x[)k( 2i

2f2

1xx

221 f

i

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The work done by the spring force on the block is negative if xf > xi.

If we set at L0 , xi = 0, then the work done by the spring force on the block for extension x is

Change in potential energy of spring = - work done by the spring force.

The work done by the spring force is to increase P.E. and to decrease K.E.

221

S xkW

221 kxchange.E.P

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Work Done on the Spring byan Applied Force

If a block is stationary before and after a displacement, the work done on the block by an applied force is negative to the work done on it by the spring force.

Therefore, for an extension x, the work done by an applied force on the block is :

The work done by the applied force equals to elastic potential energy of the spring.

The work done by the applied force is to increase the P.E. of the spring.

221

A xkW

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Sample Problem 7-7

A package of spicy Cajun pralines lies on a frictionless floor, attached to the free end of a spring in the arrangement of Fig. 7-10a. An applied force of magnitude Fa = 4.9 N would be needed to hold the package

stationary at x1 = 12 mm.

(a)  How much work does the spring force do on the package if the package is pulled rightward from x0 = 0 to

x2 = 17 mm?

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SOLUTION: 

xkFS

m/N408m10x12

N9.4

x

Fk

31

S

J059.0

)m10x17()m/N408(kxW 23212

221

s

Work done by the spring is :

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(b)  Next, the package is moved leftward to x3 = -12 mm.

How much work does the spring force do on the package during this displacement? Explain the sign of this work.

SOLUTION: 

mJ30J030.0

)m10x17()m10x12()m/N408(

kxkxW2323

21

2i2

12f2

1s

The spring force does positive work as the block moves from +17mm to its relaxed position and negative work as the block moves from the relaxed position to -12mm. The former work is larger resulting in WS being positive.

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Sample Problem 7-8

In Fig. 7-11, a cumin canister of mass m = 0.40 kg slides across a horizontal frictionless counter with speed v = 0.50 m/s. It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed?

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SOLUTION: 

We assume the spring is massless. Work done by the spring on the canister is negative. This work is :

221

S kdW

Kinetic energy change of the canister is : 2

21

if mvkk Therefore,

2212

21 mvkd

cm2.1m10x2.1

m/N750

kg40.0)s/m50.0(

k

mvd

2

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Work done by a General Variable Force

)forceiablevarbywork(rdFW

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Power

The instantaneous power P is

Power is a scalar and the SI unit is J / s which is known as the Watt (W), named after James Watt.

Note that the kilowatt-hr is a unit of work

td

WdP

s/lbft738.0s/J1W1watt1 W746s/lbft550hp1horsepower1

)s3600()W10(hkW1hourkiwowatt1 3MJ60.3J10x60.3 6

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Therefore,

td

xdcosF

td

xdcosF

td

WdP

cosvFP

vFP

The angle is between the force and velocity.

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Sample Problem 7-10

Figure 7-14 shows constant forces and acting on a box as the box slides rightward across a frictionless floor. Force is horizontal, with magnitude 2.0 N; force is angled upward by 60° to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s.

1F

1F

2F

2F

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(a)  What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant?

SOLUTION: 

0

PPP

W0.6

60cos)s/m0.3()N0.4(60cosvFP

W0.6

180cos)s/m0.3()N0.2(180cosvFP

21net

22

11

The kinetic energy of the box is not changing. The speed of the box remains at 3 m/s. The net power does not change.

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(b)  If the magnitude of is, instead, 6.0 N, what now is the net power, and is it changing?

2F

SOLUTION: 

W0.3

W0.9W0.6PPP

W0.9

60cos)s/m0.3()N0.6(60cosvFP

21net

22

There is a net rate of transfer of energy to the box. The kinetic energy of the box increases. The net power also increases.

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Homework (due Oct 18) 5P 7E 11P 17P 19P 21E 23P 31E 33P