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194 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Checkpoint 5

In three trials, a block is pushed

by a horizontal applied force

across a floor that is not friction-

less, as in Fig. 8-13a. The magni-

tudesFof the applied force and

the results of the pushing on the

block’s speed are given in the

table. In all three trials, the block is pushed through the same distance d. Rank the

three trials according to the change in the thermal energy of the block and floor that

occurs in that distance d, greatest first.

Trial F Result on Block’s Speed

a 5.0 N decreases

b 7.0 N remains constant

c 8.0 N increases

be friction and a change Ethin thermal energy of the crate

and the floor. Therefore, the system on which the work is

done is the crate – floor system, because both energy

changes occur in that system.

(b) What is the increase Ethin the thermal energy of the

crate and floor?

KEY IDEA

We can relate Ethto the work Wdone by with the energy

statement of Eq. 8-33 for a system that involves friction:

WEmecEth. (8-34)

Calculations:We know the value of W from (a). The

changeEmecin the crate’s mechanical energy is just the

change in its kinetic energy because no potential energy

changes occur, so we have

Substituting this into Eq. 8-34 and solving for Eth, we find

(Answer)

Without further experiments, we cannot say how much of

this thermal energy ends up in the crate and how much in

the floor. We simply know the total amount.

22.2 J22 J.

20 J^12 (14 kg)[(0.20 m/s)^2 (0.60 m/s)^2 ]

EthW(^12 mv^2 ^12 mv 02 )W^12 m(v^2 v 02 )

EmecK^12 mv^2 ^12 mv 02.

F

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Sample Problem 8.05 Work, friction, change in thermal energy, cabbage heads

A food shipper pushes a wood crate of cabbage heads (total

mass m 14 kg) across a concrete floor with a constant

horizontal force of magnitude 40 N. In a straight-line dis-

placement of magnitude d 0.50 m, the speed of the crate

decreases from v 0 0.60 m/s to v 0.20 m/s.

(a) How much work is done by force , and on what system

does it do the work?

KEY IDEA

Because the applied force is constant, we can calculate

the work it does by using Eq. 7-7 ( ).

Calculation:Substituting given data, including the fact that

force and displacement are in the same direction, we

find

WFdcosf(40 N)(0.50 m) cos 0

20 J. (Answer)

Reasoning:To determine the system on which the work is

done, let’s check which energies change. Because the crate’s

speed changes, there is certainly a change Kin the crate’s

kinetic energy. Is there friction between the floor and the

crate, and thus a change in thermal energy? Note that and

the crate’s velocity have the same direction. Thus, if there is

no friction, then should be accelerating the crate to a

greaterspeed. However, the crate isslowing,so there must

F

:

F

:

d

:

F

:

WFd cos 

F

:

F

:

 



F

 :

Additional examples, video, and practice available at WileyPLUS

changes, and the thermal energies of the block and floor also change. Therefore, the

work done by force is done on the block – floor system. That work is

WEmecEth (work done on system, friction involved). (8-33)

This equation, which is represented in Fig. 8-13b, is the energy statement for the

work done on a system by an external force when friction is involved.

F

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