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7-3 Newton’s second law applied to an object allows us to determine a formula for kinetic energy 301

EXAMPLE 7-3 Up the Hill Revisited
.
Consider again the box of supplies that you pushed up the incline in Example 7-2 Use the work-energy theorem for an

object to find how fast the box is moving when it reaches your campsite, assuming it is moving uphill at 0.75m/s at the

bottom of the incline.
Set Up
Our goal is to find the final speed v of the box. In Work-energy theorem for an object: v v v = ?

=
f
f f f f
Example 7-2 we were given the weight (and hence the = K − K i (7-9)

mass m ) of the box, and we found that the net work W net f v v v = 0.75 m/s
Kinetic energy:
i i

.
done on the box is W net = 56 J Because we’re given

= 0.75 m/s we can use Equations 7-8 and 7-9 to 1 W W W = 56 J
,
v i = mv (7-8) ne net
2

K
v .
solve for f 2
Solve

Use the work-energy theorem for an object to solve for v Combine Equations 7-8 and 7-9 :
f
,

in terms of the initial speed, i v the net work done, net , 1 1
W

2
and the mass m . W net = mv − mv 2 i
f
1
2

Isolate the mv term on one side of the equation, 1 2 2
1
f
2

v by multiplying through by m2/ and 2 2
then solve for f mv = mv + W net
i
f
taking the square root. 2 2

2
2
v = v + 2W net
f
i
m
2W net
2
f v = v +
i
m
g F 225 N
.
= 56 J We find m = = = 23.0kg

From Example 7-2 we know that W net 2
the mass m of the box from its known weight g F = 225 N g 9.80 m/s

and the relationship g F = mg . Use these to find the value f v = ( 0.75 m/s) + 2(56 J)

2
v
of f . 23.0kg
= 2.3m/s
Reflect
We could have solved this problem by first finding the net force on the box, then calculating its acceleration from Newton’s
.

second law, and finally solving for v by using one of the kinematic equations from Chapter 2 This would result in the
f
correct answer, but using the work-energy theorem for an object is easier and gives fewer opportunities to make a mistake.
As a check on our result, note that we found in Example 7-2 that the net force on the box is 9.4 N uphill. We can
use Equation 2-11 to find the final velocity after moving 6.00 m up the incline and it agrees, as it must if you did not
make a math mistake.
NOW WORK Problems 2, 3, and 6 from The Takeaway 7-3.
Example 7-3 llustrates how using the work-energy theorem can simplify solving
i
problems. In the following section we’ll see more examples of how to apply this power-
ful theorem to various cases of linear motion. In Section 7-5 we’ll see how the work-
energy theorem can be applied to problems in which the motion is along a curved path
and in which the forces are not constant.
THE TAKEAWAY for Section 7-3

✔ The net work done on an object (the sum of the work is equal to the amount of work that would be needed to
done on it by all forces exerted on it) as it undergoes a accelerate the object from rest to its present speed.
displacement is equal to the change in the object’s kinetic ✔ The kinetic energy of an object is also equal to the

energy during that displacement. maximum amount of work the object can do in the process

✔ The formula for the kinetic energy of an object of mass of coming to a halt from its present speed.
2
.

m and speed v is =K 1 2 mv The kinetic energy of an object
Uncorrected proofs have been used in this sample. Copyright © Bedford, Freeman & Worth Publishers.
Distributed by Bedford, Freeman & Worth Publishers. For review purposes only. Not for redistribution.
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