Page 90 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 7] ENERGY 75

The mass of the crate does not matter here. Since the force is parallel to the displacement,
W = Fs = (420 N)(5m) = 2100 J = 2.1kJ

SOLVED PROBLEM 7.2
The 420-N force of Prob. 7.1 is instead exerted on the crate at an angle of 35 above the horizontal. How
◦
much work is done now?

◦
W = Fs cos θ = (420 N)(5m)(cos 35 ) = 1920 J = 1.92 kJ
SOLVED PROBLEM 7.3

A 60-kg box is pushed 12 m across a horizontal floor by a horizontal force of 200 N. The coefficient
of kinetic friction is 0.3. (a) How much work went into overcoming friction and (b) how much into
accelerating the box?
(a) The frictional force to be overcome is
2
F f = µ k N = µ k mg = (0.3)(60 kg)(9.8 m/s ) = 176 N
The work done opposing friction is therefore

W f = F f s = (176 N)(12 m) = 2112 J
(b) The remainder F of the applied force goes into accelerating the box. Since

F = F − F f = 200 N − 176 N = 24 N
the work done in accelerating the box is

W a = ( F)s = (24 N)(12 m) = 288 J

SOLVED PROBLEM 7.4

How much work is done in raising a 2-kg book from the ground to a height of 1.8 m?
The force needed to raise the book is equal to its weight mg. This force acts parallel to the displace-
ment of the book, where s = h,so

2
W = Fs = mgh = (2kg)(9.8 m/s )(1.8m) = 35 J
SOLVED PROBLEM 7.5

How much work is done in raising a 2000-lb elevator through a height of 80 ft?
The force here equals the elevator’s weight w. Hence,
5
W = Fs = wh = (2000 lb)(80 ft) = 1.6 × 10 ft·lb

SOLVED PROBLEM 7.6
In raising a 200-kg bronze statue 10,000 J of work is performed. How high is it raised?

Since W = Fs = mgh,
4
W 1 × 10 J
h = = = 5.1m
2
mg (200 kg)(9.8 m/s )