# Homework Help: Mechanics/Kinetics of Particles

1. May 28, 2006

### scluggy

Two weights are hanging from a rope that goes through two pulleys as shown below:

_______
| O
| / \
| | |
\ / |
O |
| |
A B

A weighs 3 kg
B weighs 2 kg

B has an initial speed of 0.8 m/s

How far will B drop before A reaches a speed of 0,6 m/s?

The mass of the pulley and the cable can be neglected.

I have been trying to resolve this for some time now but will always get the wrong answer. The way I believe it should work is:

The potential energy + the kinetic energy for the system is the same at the start and end of this so we would have a conversion of energy from potential energy to kinetic energy. So I have set it up as having 0 potential energy to start off with (height = 0)

mgh = 2*9.81*0 = 0 and 3*9.81*0 = 0

The velocity for B would be twice that of the velocity for A due to the pulley system.

The initial kinetic energy would be:
0.5mv^2 = 0.5*2*0.8^2 and 0.5*3*-0.4^2

So the whole amount of energy for this would be
0.64-0.24 = 0.4

If we then set this equal to the same formula but for the A speed of 0.6 m/s we get
0.88 = (kinetic energy A) + (kineric energy B) + (potential energy A) + (potential energy B)

0.4 = 0.5*3*-0.6^2 + 0.5*2*1.2^2 + 3*9.81*h/2 + 2*9.81*h

Which makes h = - 100 / 6867

This is incorrect as the answer should be 0.224

Can someone see where I am going wrong and perhaps help with how this should be resolved? I have been trying to resolve this issue for some time now with different methods but will always come up with the wrong answer.

Thankful for any suggestions!

2. May 28, 2006

### Staff: Mentor

Your error is in having a negative kinetic energy! The KE of mass B is positive, not negative.

Regardless of your sign convention, the speed is squared so the KE is always positive. (To maintain sanity, I would choose a sign convention of up = +, down = -.)

3. May 28, 2006

### scluggy

Thank you so much!!! I was really struggling with this problem but now it's solved thanks to your help :rofl: