# Dynamics: Determining amount of energy dissapated by a pulley

1. Feb 23, 2012

### jaredogden

1. The problem statement, all variables and given/known data

Blocks A and B have masses 11 kg and 5 kg, respectively, and they are both at a height h = 2m above the ground when the system is released from rest. Just before hitting the ground block A is moving at a speed of 3 m/s. Determine a.) the amount of energy dissipated in friction by the pulley, b.) the tension in each portion of the cord during the motion.

(blocks A and B are hanging at the same height attached by a cable that passes through a pulley above them)

2. Relevant equations

KE = 1/2mv2
PE = mgz
F = ma

3. The attempt at a solution

EA = mgz
EA = (11 kg)(9.81 m/s2)(2 m)

EB = mgz
EB = (5 kg)(9.81 m/s2)(2 m)

KEA at y = 0 m
KEA = 1/2(11 kg)(3 m/s)2
KEA = 49.5 J

The answer in the book for a.) is 45.7 J

2. Feb 23, 2012

### LawrenceC

You have to look at the entire system, not only the kinetic energy in block A. What you calculated is the KE for block A. Write an energy balance for the entire system such as:

Initial energy (KE+PE) = Final energy (KE+PE) + pulley loss

You'll see you will get the correct answer.

3. Feb 23, 2012

### jaredogden

Oh thank you! I was trying that earlier but I was trying to figure out the height that block B would be at to add in its potential energy.

Just adding their original potentials and subtracting them from their kinetic energies assuming they were traveling at the same speed did the trick. Thanks again.

4. Feb 23, 2012

### LawrenceC

In order to find the tensions you should draw a free body diagram of each mass. You will see forces such as tensions, weights, and forces due to acceleration for each. Equate the forces. Each equation will have two unknowns, namely the tension and the acceleration. The acceleration is the same for each force balance.

So you now have two equations with three unknowns. Hint: Determine the acceleration by another means.

5. Feb 23, 2012

### jaredogden

Ah perfect! Once I saw the equate the forces I felt stupid.

Thanks so much for the help I finally got it!