SOL21 - Thevalingam Donald Homework 21 Due Mar 4 2007...

Thevalingam, Donald – Homework 21 – Due: Mar 4 2007, midnight – Inst: Eslami 1
This print-out should have 12 questions.
Multiple-choice 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.