Page 67 - Schaum's Outline of Theory and Problems of Applied Physics

52 LAWS OF MOTION [CHAP. 5

Since the ball’s mass is m = 60 g = 0.06 kg, the average force on it was
3
2
F = ma = (0.06 kg)(−7 × 10 m/s ) =−420 N
The minus sign means that the force was in the opposite direction to that in which the ball approached the racket.

SOLVED PROBLEM 5.7

The brakes of a 1000-kg car exert 3000 N. (a) How long will it take the car to come to a stop from a
velocity of 30 m/s? (b) How far will the car travel during this time?

(a) The acceleration the brakes can produce is
F −3000 N 2
a = = =−3 m/s
m 1000 kg
(The force is considered negative because it acts opposite to the car’s direction of motion.) Here the initial
2
velocity is v 0 = 30 m/s, the final velocity is v = 0, and the acceleration is −3 m/s . Hence
v = v 0 + at
2
0 = 30 m/s − (3 m/s )(t)
30 m/s
t = = 10 s
3 m/s 2
1
2
2
1
2
(b) s = v 0 t + at = (30 m/s)(10 s) − (3 m/s )(10 s) = 300 m − 150 m = 150 m
2 2
WEIGHT
The weight of a body is the gravitational force with which the earth attracts the body. If a person weighs 600 N
(135 lb), this means the earth pulls that person down with a force of 600 N. Weight (a vector quantity) is different
from mass (a scalar quantity), which is a measure of the response of a body to an applied force. The weight of a
body varies with its location near the earth (or other astronomical body), whereas its mass is the same everywhere
in the universe.
The weight of a body is the force that causes it to be accelerated downward with the acceleration of gravity g.
Hence, from the second law of motion, with F = w and a = g,
w = mg

Weight = (mass)(acceleration of gravity)
Because g is constant near the earth’s surface, the weight of a body there is proportional to its mass—a large
mass is heavier than a small one.

SOLVED PROBLEM 5.8

A 100-kg man slides down a rope at constant speed. (a) What minimum breaking strength must the rope
have? (b) If the rope has precisely this strength, will it support the man if he tries to climb back up?

(a) Since the man slides down at constant speed, the rope must support his weight of mg = 980 N.
(b) For the man to climb up, he must exert an additional force on the rope, so it will break.

SOLVED PROBLEM 5.9
(a) What is the weight of an object whose mass is 5 kg? (b) What is its acceleration when a net force of
100 N acts on it?