# Homework Help: Conservation of Energy of blocks and rope

1. Jul 14, 2013

### SPiazzo

Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended 1.24 m, its speed is 3.27 m/s. If the total mass of the two blocks is 16.3 kg, what is the mass of each block? (Enter your answers from smallest to largest.)

Oh and I don't think Ei=Ef will work here as if you do that then mgh+1/2MVi^2=1/2MV^2. Where 1/2MVi^2 =0 then you have mgh=1/2MV^2 and the m's cancel out. So that equation will not work!!

2. Jul 14, 2013

### BruceW

what is m and what is M ? And think about the direction of movement of the blocks. Will they both lose PE?

3. Jul 14, 2013

### SPiazzo

m and M are both mass one will lose potential energy the other will gain, but that really doesn't help me. As I can't visualize that, maybe if you helped me by putting in an equation or formula I would understand. As simply writing a one sentence response does not seem to work and is quite honestly not that helpful.

4. Jul 14, 2013

### HallsofIvy

What will work is the fact that if a mass, m, has force f applied to it, it will move, starting with 0 speed, a distance $(1/2)(f/m)t^2$ and have speed $(f/m)t$. Since we are told that . After this block has descended 1.24 m, its speed is 3.27 m/s, we have $(1/2)(f/m)t^2= 1.24$ and $(f/m)t= 3.27$. From the second equation, [itex]t= 3.27(m/f)[itex] and, putting that into the first, [itex](1/2)(f/m)(3.27)^2(m^2/f^2)= 5.34645(m/f)= 1.24 so m/f= 1.24/5.34645= 0.2319....

Of course, there is a force mg on the heavier mass and a force of Mg on the smaller mass which, through the pully becomes an upward force on the heavier mass. That is, the total force on the heavier mass is f= mg- Mg= (m- M)g. So the previous equation becomes m/((m-M)g)= 0.2319. g= 9.81, approximately, so m/(m- M)= g(.2319)= 2.725.... That is equivalent to m= (m- M)(2.725)= 2.725m- 2.725M or 2.725M= 2.725m- m= 1.725M. That, together with m+ M= 16.3, gives two equations to solve for m and M.

5. Jul 14, 2013

### BruceW

I can't really write out the equation without just doing the problem. This forum is definitely not for giving direct solutions to problems, it is more for helping/advice. Right, so starting with your equation: mgh+1/2MVi^2=1/2MV^2 You have only included KE due to one of the masses. what about the other one? and you have only included the change in potential energy of one of the masses. but both masses have a change in potential energy. try writing out the equation. you also know that m + M = Mtotal so it is essentially a simultaneous equation problem.

edit: p.s. it is possible to get the answer using just conservation of energy and the m + M = Mtotal equation. Don't give up too early on it.

edit again: ah wait, you were using m and M to both mean the total mass? well first, try using m1 and m2 (for the masses of the individual blocks) and use them in the energy equation. and think about the direction of movement of the two blocks. Do they both lose PE ?

Last edited: Jul 14, 2013