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7-3 Newton’s second law applied to an object allows us to determine a formula for kinetic energy 297

(d) On the horizontal segment ii. you exert a force on Hero of Alexandria wrote about what we now call
the box as you walk but the kinetic energy of the box simple machines in the first century, c.e .
does not change. On this path the box exerts an equal “Whenever we want to move some weight, if we tie a rope
and opposite force on you but there is no work done to this weight we pull with as much force as is equal to the
by that force on you. If you walked 3 km rather than burden. But if we untie the rope from the weight, and tie one
3 m you would probably notice that you had used a of its ends to a stationary point and pass its other end over a
lot more of your internal energy. Many students claim pulley fastened to the burden and draw on the rope, we will
that the extra energy you used is transferred to the more easily move the weight.”
box. Evaluate this claim and justify your answer. In Hero’s machine a rope runs over a pulley connected to
(e) How would your answer to part (c) change if you did a stationary beam (A), then runs through a second pulley

not neglect when you were changing the velocity of (B) that is attached to the “burden,”
the box? then runs back to the beam where it A
is tied as shown in the diagram.

Skills in Action (b) This machine is used to lift the
burden to the same height, ∆x.

8. Some technologies are ancient. One example Predict the work done on the B

is the pulley. To lift a burden above the floor, burden, the external force that m
a pulley is attached to the ceiling and a rope must be exerted to do this work, F
runs over the pulley and is attached to the and the displacement of the free
burden (as shown in the figure). A force F is F end of the rope as the burden is raised. Assume that
exerted on the rope and the burden is then pulley B has a negligible mass.
displaced upward by ∆x at a constant speed. (c) Justify the claim that this technology reduces the

(a) Express the work done on the burden by the external effort required to achieve the same outcome required
force in terms of F and ∆x. to lift the burden to the same height.
Newton’s second law applied to an object allows us to 1 A cart of mass m moves
with negligible friction
7-3 determine a formula for kinetic energy and state the along a line that we’ll
call the x axis.
work-energy theorem for an object
d
In Example 7-2 we considered a box being pushed uphill that gained speed as it moved. We

found that the total amount of work being done on this box was positive. Let’s now show F
that there’s a general relationship between the total amount of work done on an objec t and
the change in that object’s speed. To find this important relationship, which defines kinetic
energy, we’ll combine our definition of work from Section 7-2 with what we learned about x x x
.

one-dimensional motion in Chapter 2 and our knowledge of Newton’s laws from Chapter 4 i f
Remember that for something that can be modeled as an object, all points on that object
must move the same distance in any motion, so, the point of contact of a force exerted on 2 The cart moves from initial
an object and the center of mass of the object always have the same displacement. position x to nal position x . f
i
Let’s begin by considering an object that moves along a line that we’ll call the

x axis, traveling from initial position x i to final position x with a constant acceleration 3 As the cart moves, a constant

f

a x ( Figure 7-8 ). net force in the x direction is

,
,
Equation 2-11 reintroduced here as Equation 7-3 gives us a relation between the exerted on the cart.

object’s velocity i v x at i x and its velocity v at :x Figure 7-8 Deriving the work-

f
fx
2
v 2 = v 2 + a x − x ) i (7-3)

(
x f x i x f energy theorem for an object
2
In Equation 7-3 v is the square of the velocity at x , but it also equals the square of A cart moves along a line, traveling

f
fx

v at x That’s because v is equal to +v if the object is moving in the from initial position i x to final
.

the object’s speed f f fx f position f x . The net force exerted on

positive x direction and equal to −v if the object is moving in the negative x direction. the cart is constant, so the cart has a

f
2
2
,

In either case, v 2 fx = v For the same reason v 2 ix = v where i v is the object’s speed at i x . constant acceleration a x .

.

i
f
So we can rewrite Equation 7-3 as
Speed at position x of an object in linear Speed at position x of the object
f
i
motion with constant acceleration
EQUATION IN WORDS
Relating speed, acceleration, and
2
2
v = v + 2a (x − x ) ( 7-4 ) position for linear motion with
f
f
i
x
i
constant acceleration
Constant acceleration of the object Two positions of the 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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