9/14/23, 12:58 AM
Work
9.2.1 Work
Now let’s turn to the integral on the right-hand side of Equation 9.6 . This integral is telling us by how much the kinetic energy changes due to the force.
That is, it is the energy transferred to or from the system by the force. Earlier we said that a process that transfers energy to or from a system by mechani⃗
cal means—by forces—is called work. So the integral on the right-hand side of Equation 9.6  must be the work W done by force 𝐹.
Having identified the left side of Equation 9.6  with the changing kinetic energy of the system and the right side with the work done on the system, we
can rewrite Equation 9.6  as
(9.9)
This is our first version of the energy principle. Notice that it’s a cause-and-effect statement: The work done on a one-particle system causes the
system’s kinetic energy to change.
OT DISTR
N
I
O
We’ll study work thoroughly in the next section, but for now we’re considering only the simplest case of a constant force parallel to the direction of mo-
-
𝑊
𝑠f
𝑠f
𝑠i
𝑠i
𝑠f
= ∫ 𝐹𝑠 𝑑𝑠 = 𝐹𝑠 ∫ 𝑑𝑠 = 𝐹𝑠 𝑠| = 𝐹𝑠 (𝑠f − 𝑠i )
E
© P E A RS O N
(9.10)
B
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D
tion (the s-axis). A constant force can be factored out of the integral, giving
𝑠i
= 𝐹𝑠 Δ𝑠
The unit of work, that of force multiplied by distance, is the N m. Recall that 1 N = 1 kg m/s2 . Thus
A
1 N m = 1 ( kg m/s2 ) m = 1 kg m2 /s2 = 1 J
UN
FO
R
T
HE
2@
work in joules.
B.C
Thus the unit of work is really the unit of energy. This is consistent with the idea that work is a transfer of energy. Rather than use N m we will measure
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6G
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IVAT
E USE OF
K
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9/14/23, 12:58 AM
Work
EXAMPLE 9.1 Firing a cannonball
A 5.0 kg cannonball is fired straight up at 35 m/s. What is its speed after rising 45 m?
MODEL Let the system consist of only the cannonball, which we model as a particle. Assume that air resistance is negligible.
VISUALIZE FIGURE 9.3  is a before-and-after pictorial representation. Notice two things: First, for problem solving we use numerical subscripts
instead of the generic i and f. Second, the v symbols are speeds, not velocities, so there’s no x or y in the subscripts. Before-and-after representations are usually simpler than the pictorial representation used in dynamics problems, so you can include known information right on the diagram
instead of making a Known table.
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FIGURE 9.3 Before-and-after representation of the cannonball.
SOLVE Using work and energy is not the only way to solve this problem. You could solve it as a free-fall problem. Or you might have previously
learned to solve problems like this using potential energy, a topic we’ll take up in the next chapter. But using work and energy emphasizes how
these two key ideas are related, and it gives us a simple example of the problem-solving process before we get to more complex problems. The energy principle is Δ𝐾 = 𝑊, where work is done by the force of gravity. The cannonball is rising, so its displacement Δ𝑦 = 𝑦1 − 𝑦0 is positive. But the
A
force vector points down, with component 𝐹𝑦 = − 𝑚𝑔. Thus gravity does work
B.C
2
𝑊 = 𝐹𝑦 Δ𝑦 = − 𝑚𝑔 Δ𝑦 = − (5.0 kg)(9.80 m/s )(45 m) = − 2210 J
R
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HE
1
1
2
𝐾0 = 2 𝑚𝑣0 2 = 2 (5.0 kg)(35 m/s) = 3060 J
2@
FO
The cannonball’s change of kinetic energy is Δ𝐾 = 𝐾1 − 𝐾0 . The initial kinetic energy is
UN
as the cannonball rises 45 m. A negative work means that the system is losing energy, which is what we expect as the cannonball slows.
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K
Using the energy principle, we find the final kinetic energy to be 𝐾1 = 𝐾0 + 𝑊 = 3060 J − 2210 J = 850 J. Then
𝑣1 = √
2𝐾1
2(850 J)
=√
= 18 m/s
𝑚
5.0 kg
REVIEW 35 m/s ≈ 70 mph. A cannonball fired upward at that speed is going to go fairly high. To have lost half its speed at a height of 45 m ≈ 150 ft
seems reasonable.
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