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254 CHAPTER 8 Conservation of Energy

by the maximum force exerted by the engine on the wheels, and this is not directly
related to the power as defined above.
The SI unit of power is the watt (W), which is the rate of work of one joule per
second:

1 watt 1 W 1 J s
In engineering practice, power is often measured in horsepower (hp) units, where

1 horsepower 1 hp 746 W (8.35)
This is roughly the rate at which a (very strong) horse can do work.
Note that multiplication of a unit of power by a unit of time gives a unit of energy.
An example of this is the kilowatt-hour (kW
h), already mentioned in Section 8.3:

1 kilowatt-hour 1 kW
h 1 kW 1 h 1000 W 3600s
(8.36)
6
3.6 10 J
JAMES WATT (1736–1819) Scottish This unit is commonly used to measure the electric energy delivered to homes and
inventor and engineer. He modified and factories.
improved an earlier steam engine and founded For a constant (or average) power P delivered to a body during a time
t, the work
the first factory constructing steam engines.

W delivered is the rate times the time [see Eq. (8.33)]:
Watt introduced the horsepower as a unit of
mechanical power.
W P
t (8.37)
If the rate of doing work P varies with time, then the total work W done between a
time t and another time t is the sum of the infinitesimal P
t contributions; that is,
1 2
the work done is the integral of the power over time:
t 2
W dW P dt (8.38)

t 1

An elevator cage has a mass of 1000 kg. How many horse-
EXAMPLE 8
power must the motor deliver to the elevator if it is to raise
the elevator cage at the rate of 2.0 m/s? The elevator has no counterweight
(see Fig. 8.14).
Motor steadily
2
does work. SOLUTION: The weight of the elevator is w mg 1000 kg 9.81 m/s
9800 N. By means of the elevator cable, the motor must exert an upward force
equal to the weight to raise the elevator at a steady speed. If the elevator moves
up a distance
y, the work done by the force is

W F
y (8.39)
motor
To obtain the power,or the rate of work,we must divide this by the time interval
t:
Elevator ascends with F ¢y ¢y
constant velocity. P ¢W Fv (8.40)
¢t ¢t F ¢t

where v
y/
t is the speed of the elevator. With F 9800 N and v 2.0 m/s,
we find
4
P Fv 9800 N 2.0 m/s 2.0 10 W
Since 1 hp 746 W [see Eq. (8.35)], this equals

1000 kg 1 hp
4
P 2.0 10 W 27 hp
FIGURE 8.14 Elevator cage and motor. 746 W