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7-2 The work done by a constant force exerted on a moving object 291

EXAMPLE 7-1 Lifting a Book
How much work must you do to lift a textbook with a mass of 2.00 kg a vertical distance of 5.00 cm? You lift the book
at a constant speed.

Set Up F exerted by you
Newton’s first law tells us that the net force on the book Work done on an object, force in
must be zero if it is to move upward at a constant speed. the same direction as displacement: d = 5.00 cm
Hence the upward force you exert must be constant and
equal in magnitude to the gravitational force on the book. W = Fd (7-1)
Because the force you exert is constant and in the direction book, m = 2.00 kg
in which the book moves, we can use Equation 7-1 to
calculate the work done on the book by that force.

F = mg
g
Solve
2
Calculate the magnitude F of the force that you exert on the F = mg = (2.00kg)(9.80m/s )
book, then substitute this and the displacement = 5.00 cmd = 19.6N
into Equation 7-1 to determine the work that you do on
the book. W = Fd = (19.6 N)(5.00cm)
= (19.6 N)(0.0500 m)
= 0.980 Nm
= 0.980 J

Reflect

The work that you do in lifting the book is almost exactly 1 joule. We don’t yet have much practice with units of energy
and work, but consider the basic definition of a joule. You exerted nearly 20 N of force on the book, but only for 5 cm
of motion, 1/20th of a meter. The product then should be about 1 Nm, a little lower because the force you exerted wasn’t
quite 20 N. That is what we got! You may be worried because you did work on the book and did not change its kinetic
energy and we said that work was a way to transfer energy. We will find that this was because your lift was not the only
force exerted on the book. The force of gravity was pulling exactly opposite you. We will learn how to calculate the
work done by a force in the direction opposite to displacement after the next example.
Extend: The actual amount of energy that you would need to expend is several times more than 0.980 J. That’s because
your body isn’t 100% efficient at converting energy into work. Some of the energy goes into lifting your arm, and some
into heating your muscles. In fact, your muscles can consume energy even when they do no work, as we describe below.
The value of 0.980 J that we calculated is just the amount of work that you do on the book.

NOW WORK Problem 1 from The Takeaway 7-2.

Muscles and Doing Work

Pick up a heavy object and hold it in your hand at arm’s length (Figure 7-3). After
a while you’ll notice your arm getting tired: It feels like you’re doing work to hold
the object in midair. But Equation 7-1 says that you’re doing no work on the object
because it isn’t moving (its displacement d is zero). So why does your arm feel tired?
The reason your arm tires while holding a heavy object is that whether you are
moving the object or not, you still have to exert a force to balance that of gravity (it
is the same size force, and in the opposite direction) to hold the object at a constant
height. Your body has to convert energy stored in the chemical bonds of fat and sugar
(chemical energy) to a type that your muscle cells can use to exert this force, but these Figure 7-3 Getting tired while
reactions generate waste products that change the conditions in the muscle, including doing zero work Weights that you
hold stationary in your outstretched
warming it up. This process of changing energy from one type into another leads to the arms undergo no displacement, so you
feeling that you’ve been doing work—even though you’re doing no work on the object do zero work on them. Why, then, do
you’re holding. your arms get tired?

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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