Thevalingam, Donald – Homework 21 – Due: Mar 4 2007, midnight – Inst: Eslami
1
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printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A 36 kg child on a 2
.
3 m long swing is released
from rest when the swing supports make an
angle of 25
◦
with the vertical.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If the speed of the child at the lowest po
sition is 1
.
83818 m
/
s, what is the mechani
cal energy dissipated by the various resistive
forces (
e.g.
friction, etc.)?
Correct answer: 15
.
2051 J.
Explanation:
The total energy at the top of the swing is
potential, and the total energy at the bottom
is kinetic.
At the top of the swing, the swing is a
‘
cos
θ
below the support, so it is
‘
(1

cos
θ
) above
its lowest point. The mechanical energy lost
due to friction would thus be
Δ
E
=
E
top

E
bottom
=
mgh

1
2
mv
2
= (36 kg)(9
.
8 m
/
s
2
)(0
.
215492 m)

1
2
(36 kg)(1
.
83818 m
/
s)
2
= 15
.
2051 J
keywords:
002
(part 1 of 3) 10 points
The coefficient of friction between the 6
.
1 kg
mass and the table is 0
.
55, and the coefficient
of friction between the 5 kg mass and the table
is 0
.
3.
Consider the motion of the 39 kg mass
which descends by the amount of Δ
y
= 0
.
3 m
after releasing the system from rest.
The acceleration of gravity is 9
.
8 m
/
s
2
.
6
.
1 kg
μ
= 0
.
55
5 kg
μ
= 0
.
3
39 kg
T
12
T
13
a
g
Find the work done against friction.
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 Spring '08
 Turner
 Physics, Energy, Kinetic Energy, Mass, Work, kg, Thevalingam

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