A bucket of water with a mass of 2.0 kg is attached to a rope that is wound arou
ID: 1264613 • Letter: A
Question
A bucket of water with a mass of 2.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 8.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest.
(a) Find its speed after it has fallen through a distance of 1.30 m?
m/s
(b) What is the tension in the rope?
N
(c) What is the acceleration of the bucket?
m/s2
Explanation / Answer
The kinetic energy gained by the bucket
1/2mv^2 and the kinetic energy gained by the cylinder is
1/2 I v^2/r^2.
I = mr^2/2 and hence
1/2 I v^2/r^2 = 1/2 *mr^2/2 *v^2/r^2 = 1/4 mv^2.
Total energy = 1/2 *4*v^2 + 1/4*2*v^2 = 2.5 v^2
Potential energy lost = mgh = 2*9.8*1.30=25.48.
2.5 v^2 = 2*9.8*1.3
v = 3.19 m/s
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b)
The acceleration of the bucket is given by
v^2 = 2as.
a = v^2/2s =3.19^2 / [2*1.3]
= 3.914 m/s^2.
Tension T = ma - mg = m [a-g]
2* [3.914 -9.81] = -11.792 N
Minus sign since it is upward.
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c)
a = 3.914 m/s^2.
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