Page 93 - Schaum's Outline of Theory and Problems of Applied Physics

78 ENERGY [CHAP. 7

The force supplied by the car’s engine must equal this amount. The car’s velocity is
ft/s

v = (30 mi/h) 1.47 = 44 ft/s
mi/h
and so, from P = Fv, the required power at 100 percent efficiency is
4
P output = Fv = (417 lb)(44 ft/s) = 1.835 × 10 ft·lb/s
Since 1 hp = 550 ft·lb/s,
4
1.835 × 10 ft·lb/s
P output = = 33.4hp
550 (ft·lb/s)/hp
Because efficiency = P output /P input , at 70 percent efficiency,
33.4hp
P input = = 47.7hp
0.70

Fig. 7-2

ENERGY

Energy is that property something has which enables it to do work. The more energy something has, the more
work it can perform. Every kind of energy falls into one of three general categories: kinetic energy, potential
energy, and rest energy.
The units of energy are the same as those of work, namely, the joule and the foot-pound.

KINETIC ENERGY

The energy a body has by virtue of its motion is called kinetic energy. If the body’s mass is m and its velocity is
v, its kinetic energy is
1
Kinetic energy = KE = mv 2
2
SOLVED PROBLEM 7.14

Find the energy of a 1000-kg car whose velocity is 20 m/s.

1
1 2 2 5
KE = mv = (1000 kg)(20 m/s) = 2 × 10 J
2 2
SOLVED PROBLEM 7.15
What velocity does a 1-kg object have when its kinetic energy is 1 J?
1
2
2
2
2
Since KE = mv and1J = 1N·m = 1(kg·m/s ) (m) = 1kg·m /s ,
2

2(KE) 2(1J) √
v = = = 2 m/s = 1.4 m/s
m 1kg