9781118230725.pdf

Work and Kinetic Energy

Finding an Expression for Work

Let us find an expression for work by considering a bead that can slide along

a frictionless wire that is stretched along a horizontal xaxis (Fig. 7-2). A constant

force , directed at an angle fto the wire, accelerates the bead along the wire.

We can relate the force and the acceleration with Newton’s second law, written

for components along the xaxis:

Fxmax, (7-3)

wheremis the bead’s mass. As the bead moves through a displacement , the

force changes the bead’s velocity from an initial value to some other value.

Because the force is constant, we know that the acceleration is also constant.

Thus, we can use Eq. 2-16 to write, for components along the xaxis,

(7-4)

Solving this equation for ax, substituting into Eq. 7-3, and rearranging then give us

(7-5)

The first term is the kinetic energy Kfof the bead at the end of the displacement

d, and the second term is the kinetic energy Kiof the bead at the start. Thus, the

left side of Eq. 7-5 tells us the kinetic energy has been changed by the force, and

the right side tells us the change is equal to Fxd. Therefore, the work Wdone on

the bead by the force (the energy transfer due to the force) is

WFxd. (7-6)

If we know values for Fxandd, we can use this equation to calculate the work W.

1

2 mv

(^2) ^1
2 mv^0
(^2) F
xd.
v^2 v 02  2 axd.
v: 0 v:
d
:

F

:

152 CHAPTER 7 KINETIC ENERGY AND WORK

To calculate the work a force does on an object as the object moves through some

displacement, we use only the force component along the object’s displacement.

The force component perpendicular to the displacement does zero work.

Figure 7-2A constant force directed at

anglefto the displacement of a bead

on a wire accelerates the bead along the

wire, changing the velocity of the bead

from to. A “kinetic energy gauge”

indicates the resulting change in the kinet-

ic energy of the bead, from the value Kito

the value Kf.

InWileyPLUS, this figure is available as

an animation with voiceover.

:v 0 v:

d

F:

:

A

From Fig. 7-2, we see that we can write FxasFcosf, where fis the angle

between the directions of the displacement and the force. Thus,

WFdcosf (work done by a constant force). (7-7)

F

:

d

:

x x

Bead

φ Wire

F

Ki

Kf

v

v 0

This component

does no work.

Small initial

kinetic energy

Larger final

kinetic energy

This force does positive work

on the bead, increasing speed

and kinetic energy.

This component

does work.

φ

F

φ

F

φ

F

Displacementd