A bucket of water with a mass of 7.0 kg is attached to a rope that is wound arou
ID: 2259067 • Letter: A
Question
A bucket of water with a mass of 7.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 6.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest.
A bucket of water with a mass of 7.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 6.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest. Find its speed after it has fallen through a distance of 0.80 m? What is the tension in the rope? What is the acceleration of the bucket?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 *7*v^2 + 1/4*6*v^2 = 5 v^2
Potential energy lost = mgh = 7*9.8*0.80.
=54.88
5 v^2 = 54.88
v = 3.31 m/s
c))The acceleration of the bucket is given by
v^2 = 2as.
a = v^2/2s =3.31^2 / [2*0.80]
= 6.84 m/s^2.
Tension T = ma - mg = m [a-g]
7* [6.84 -9.81] = -20.66 N
Minus sign since it is upward.
b)
a = 20.66 m/s^2.
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