Page 97 - Schaum's Outline of Theory and Problems of Applied Physics

82 ENERGY [CHAP. 7

SOLVED PROBLEM 7.24
At her highest point, a girl on a swing is 7 ft above the ground, and at her lowest point she is 3 ft above
the ground. What is her maximum velocity?
The girl’s maximum velocity v occurs at the lowest point. Her kinetic energy there equals her loss of potential
energy in descending through a height of h = 7ft − 3ft = 4 ft. Hence

KE = PE
1 mv = mgh
2
2
√

2

v = 2gh = (2)(32 ft/s )(4ft) = 256 ft/s = 16 ft/s
This result is independent of the girl’s mass.
SOLVED PROBLEM 7.25
A man skis down a slope 200 m high. If his velocity at the bottom of the slope is 20 m/s, what percentage
of his initial potential energy was lost due to friction and air resistance?

Final KE 1 2 mv 2 v 2 (20 m/s) 2
= = = = 0.102 = 10.2%
2
Initial PE mgh 2gh (2)(9.8 m/s )(200 m)
which means 89.8 percent of the intial PE was lost.

SOLVED PROBLEM 7.26
A 30-kg crate is pulled up a ramp 15 m long and 2 m high by a constant force of 100 N. The crate starts
from rest and has a velocity of 2 m/s when it reaches the top of the ramp. What is the frictional force
between the crate and the ramp?
According to conservation of energy,

W = KE + PE + W f
Work done by applied change in KE change in PE work done against
= + +
force on crate of crate of crate friction
Since the length of the ramp is s = 15 m and its height is h = 2m,

W = Fs = (100 N)(15 m) = 1500 J
1
1 2 2
KE = KE final − KE initial = mv = (30 kg)(2 m/s) = 60 J
2 2
2
PE = PE final − PE initial = mgh = (30 kg)(9.8 m/s )(2m) = 588 J
The work done against friction is therefore

W f = W − KE − PE = 1500 J − 60 J − 588 J = 852 J

If F f is the frictional force, then W f = F f s and

W f 852 J
F f = = = 56.8N
s 15 m