A contractor wishes to lift a load of bricks from the sidewalk to the top of a building by means of a winch. We can now calculate how much work the force applied by the winch must do on the load to make the lift. The contractor, however, is much more interested in the rate at which that work is done. Will the job take 5 minutes (acceptable) or a week (unacceptable)?
The time rate at which work is done by a force is said to be the power due to the force. If a force does an amount of work W in an amount of time Δt, the average power due to the force during that time interval is
The instantaneous power P is the instantaneous time rate of doing work, which we can write as
Suppose we know the work W(t) done by a force as a function of time. Then to get the instantaneous power P at, say, time t = 3.0 s during the work, we would first take the time derivative of W(t) and then evaluate the result for t = 3.0 s.
The SI unit of power is the joule per second. This unit is used so often that it has a special name, the watt (W), after James Watt, who greatly improved the rate at which steam engines could do work. In the British system, the unit of power is the foot-pound per second. Often the horsepower is used. Some relations among these units are
Inspection of Eq. 7-42 shows that work can be expressed as power multiplied by time, as in the common unit kilowatt-hour. Thus,
Fig. 7-15 The power due to the truck’s applied force on the trailing load is the rate at which that force does work on the load.
Perhaps because they appear on our utility bills, the watt and the kilowatt-hour have become identified as electrical units. They can be used equally well as units for other examples of power and energy. Thus, if you pick up a book from the floor and put it on a tabletop, you are free to report the work that you have done as, say, 4 × 10−6 kW · h (or more conveniently as 4 mW · h).
We can also express the rate at which a force does work on a particle (or particle-like object) in terms of that force and the particle’s velocity. For a particle that is moving along a straight line (say, an x axis) and is acted on by a constant force 
directed at some angle 
to that line, Eq. 7-43 becomes
Reorganizing the right side of Eq. 7-47 as the dot product · 
, we may also write the equation as
For example, the truck in Fig. 7-15 exerts a force on the trailing load, which has velocity at some instant. The instantaneous power due to is the rate at which does work on the load at that instant and is given by Eqs. 7-47 and 7-48. Saying that this power is “the power of the truck” is often acceptable, but we should keep in mind what is meant: Power is the rate at which the applied force does work.
CHECK POINT 5 A block moves with uniform circular motion because a cord tied to the block is anchored at the center of a circle. Is the power due to the force on the block from the cord positive, negative, or zero?
Figure 7-16 shows constant forces 
and 
acting on a box as the box slides rightward across a frictionless floor. Force is horizontal, with magnitude 2.0 N; force is angled up ward by 60° to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s.
(a) What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant?
Solution: A Key Idea here is that we want an instantaneous power, not an average power over a time period. Also, we know the box’s velocity (rather than the work done on it).
Fig. 7-16 Two forces and act on a box that slides rightward across a frictionless floor. The velocity of the box is .
Therefore, we use Eq. 7-47 for each force. For force , at angle 
= 180° to velocity , we have
This negative result tells us that force is transferring energy from the box at the rate of 6.0 J/s.
For force , at angle 
= 60° to velocity , we have
This positive result tells us that force is transferring energy to the box at the rate of 6.0 J/s.
A second Key Idea is that the net power is the sum of the individual powers:
which tells us that the net rate of transfer of energy to or from the box is zero. Thus, the kinetic energy (K = 
mv2) of the box is not changing, and so the speed of the box will remain at 3.0 m/s. With neither the forces and nor the velocity changing, we see from Eq. 7-48 that P1 and P2 are constant and thus so is Pnet.
(b) If the magnitude of is 6.0 N, what is the net power, and is it changing?
Solution: The same Key Ideas as above give us, for the power due to ,
The power of force is still P1 = −6.0 W, and so the net power is now
which tells us that the net rate of transfer of energy to the box has a positive value. Thus, the kinetic energy of the box is increasing and so also is the speed of the box. With the speed increasing, we see from Eq. 7-48 that the values of P1 and P2, and thus also of Pnet, will be changing. Hence, Pnet = 3.0 W only at the instant the speed is 3.0 m/s.
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