9781118230725.pdf

162 CHAPTER 7 KINETIC ENERGY AND WORK

Work Done by a General Variable Force

One-Dimensional Analysis

Let us return to the situation of Fig. 7-2 but now consider the force to be in the

positive direction of the xaxis and the force magnitude to vary with position x.

Thus, as the bead (particle) moves, the magnitude F(x) of the force doing work on

it changes. Only the magnitude of this variable force changes, not its direction,

and the magnitude at any position does not change with time.

7-5WORK DONE BY A GENERAL VARIABLE FORCE

After reading this module, you should be able to...

7.14Given a variable force as a function of position, calculate

the work done by it on an object by integrating the function

from the initial to the final position of the object, in one or

more dimensions.

7.15Given a graph of force versus position, calculate the

work done by graphically integrating from the initial

position to the final position of the object.

7.16Convert a graph of acceleration versus position to a

graph of force versus position.

7. 1 7Apply the work–kinetic energy theorem to situations
   where an object is moved by a variable force.

Learning Objectives

●When the force on a particle-like object depends on
the position of the object, the work done by on the ob-
ject while the object moves from an initial position riwith
coordinates(xi,yi,zi)to a final position rfwith coordinates
(xf,yf,zf)must be found by integrating the force. If we as-
sume that component Fxmay depend on xbut not on yor
z, component Fymay depend on ybut not on xorz, and
componentFzmay depend on zbut not on xory, then the

F

F :

: work is

●If has only an xcomponent, then this reduces to

W

xf

xi

F(x)dx.

F

:

W

xf

xi

Fxdx

yf

yi

Fydy

zf

zi

Fzdz.

Key Ideas

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It then runs into and compresses a spring of spring constant
k750 N/m. When the canister is momentarily stopped by
the spring, by what distance dis the spring compressed?

KEY IDEAS

1. The work Wsdone on the canister by the spring force is
   related to the requested distance dby Eq. 7-26 (Ws
   , with dreplacingx.


2. The work Wsis also related to the kinetic energy of the
   canister by Eq. 7-10 (KfKiW).


3. The canister’s kinetic energy has an initial value of K
   and a value of zero when the canister is momen-
   tarily at rest.

1

2 mv

2

^12 kx^2 )

Calculations:Putting the first two of these ideas together,

we write the work – kinetic energy theorem for the canister as

Substituting according to the third key idea gives us this

expression:

Simplifying, solving for d, and substituting known data then

give us

1.2 10 ^2 m1.2 cm. (Answer)

dv

A

m

k

(0.50 m/s)

A

0.40 kg

750 N/m

0 ^12 mv^2 ^12 kd^2.

KfKi^12 kd^2.