Page 66 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 5] LAWS OF MOTION 51

replaced with its equivalent in terms of the meter, the second, and the kilogram. This equivalent can be found
2
2
from F = ma:1 N = (1kg)(1 m/s ) = 1kg·m/s . A newton is about 0.2248 lb, a little less than 1 lb.
4
SOLVED PROBLEM 5.1
2
A 10-kg body has an acceleration of 5 m/s . What is the net force acting on it?
2
F = ma = (10 kg)(5 m/s ) = 50 N

SOLVED PROBLEM 5.2
2
A force of 80 N gives an object of unknown mass an acceleration of 20 m/s . What is its mass?
F 80 N
m = = = 4kg
a 20 m/s 2

SOLVED PROBLEM 5.3
A force of 3000 N is applied to a 1500-kg car at rest. (a) What is its acceleration? (b) What will its
velocity be 5 s later?
F 3000 N
(a) a = = = 2 m/s 2
m 1500 kg
2
(b) v = at = (2 m/s )(5s) = 110 m/s

SOLVED PROBLEM 5.4
2
An empty truck whose mass is 2000 kg has a maximum acceleration of 1 m/s . What will its maximum
acceleration be when it carries a load of 1000 kg?
The maximum force available is
2
F = ma = (2000 kg)(1 m/s ) = 2000 N
When this force is applied to the total mass of the loaded truck, which is 3000 kg, the resulting acceleration will be
F 2000 N 2 2
a = = = m/s
m 3000 kg 3

SOLVED PROBLEM 5.5
A 1000-kg car goes from 10 to 20 m/s in 5 s. What force is acting on it?

v − v 0 20 m/s − 10 m/s 2
a = = = 2 m/s
t 5s
2
F = ma = (1000 kg)(2 m/s ) = 2000 N

SOLVED PROBLEM 5.6
A 60-g tennis ball approaches a racket at 15 m/s, is in contact with the racket for 0.005 s, and then
rebounds at 20 m/s. Find the average force that the racket exerted on the ball.
The final velocity of the tennis ball was v =−20 m/s since it reversed direction when it was struck by the
racket. Hence the ball’s acceleration was
v − v 0 −20 m/s − 15 m/s −35 m/s 3
a = = = =−7 × 10 m/s
t 0.005 s 0.005 s