Can someone check my work - Problem involving pulley

1. Oct 17, 2005

cgotu2

Here is the problem:

The system shown in Figure P8.20 consists of a light inextensible cord, light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is h. (please see the attached picture)

I used the conservation of energy and said that (KE + PE)final = (KE + PE)initial. Then i said there was no initial KE because the system was at rest and said that there was no final PE, because it had all been converted into KE. so i got the following expression:

mgh + mg(h/2) = 1/2mv^2 + 1/2m(v/2)^2

i solved for v and got sqrt(2.4gh) but this is wrong
can someone tell me where my error is??

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2. Oct 17, 2005

Integral

Staff Emeritus
There is no requirement for the PE=0 or for all of the energy to be KE. You need an expression giving the velocity when the blocks are a distance h apart. Since h is not specified it could be small or large.

3. Oct 17, 2005

cgotu2

okay i tried that, but i am confused as to what to call the height in the initial PE. the only variables to be used are v,h and g. I could call it h, but that would mean it was the same h as the separation.

this is what i have: (keep in mind that i do not know what to call the beginning h in the PE)

(PE of A + PE of B + KE of A + KE of B)initial = (PE of A + PE of B + KE of A + KE of B)final
because the blocks are at rest in the beginning i got the KE of A and KE of B to be zero...my expression then becomes:

mg(h?) + mg(h?) = mgh + mg(h/2) + 1/2mv^2 + 1/2m(v/2)^2

4. Oct 19, 2005

Well, your last entry is on the right track, but there are a few points you need to consider...

1) For gravitational potential energy (GPE), you can SET the level of zero GPE. Once you have done that, the GPE at other levels will be relative to this level.

2) Instead of considering Conservation of Energy as "Sum of final energy = Sum of initial energy" (assuming no dissipative forces), why not think about it in terms of gain/ loss?
For this instance, it would be Gain in KE of A + Gain in KE of B + Gain in GPE of B = Loss in GPE of A. Do you know why block B should move up and block A should move down when the system is released?

3) Note that blocks A and B are connected by a cord, and if we assume the cord never becomes slack throughout the motion, then the speed of both blocks will be the same throughout the motion.

4) Finally, this point requires a little imagination. If block A has dropped by a height of x, what will be the height block B has risen? Knowing this, what are the distances each block has moved for their separation to eventually become h?

The problem may not be as simple as it sounds, but this should increase your determination in solving it and boost your level of satisfaction when you have finally solved it!

Last edited: Oct 19, 2005
5. Oct 19, 2005

Staff: Mentor

You can take any position as your "zero PE" position. All that matters is the change in PE. So, why not take the initial position of the blocks to be where PE = 0? Thus you can define their initial postion to be at y = 0.
This is good.
Instead of using "h" as a variable (which will get real confusing since "h" is also the value of their final separation), use "y".

Your equation correctly shows that mass A moves twice as fast as mass B. But also realize that when mass B rises by amount X, mass A lowers by amount 2X. (Your equation has that incorrect.) Use this fact to determine how much each block must move to have their separation equal to h. (Find their final y positions in terms of h.) Rewrite your energy equation correctly and you'll be able to solve for the speed.

6. Oct 19, 2005