What Is the Speed of the Blocks When They Are at the Same Height?

[DrWillVKN]

Homework Statement

A system has two blocks connected by frictionless pulleys. One block (m = 3kg) is above the other (m = 2kg) vertically, as the other is lowered by a weight. Initially, they are 5m apart, vertically. When the weight is removed, what is the speed of the blocks when they are at the same height?

Homework Equations

Wext = dEmech + Wnc
Emech = U + K
U = mgh
K = 1/2 * mv^2

The Attempt at a Solution

No external forces are done on the system, and no nonconservative forces are done either. Thus, Uf + Kf = Ui + Kf. Kf = 0, and Ui = mgh for both blocks. If the height is 0 when the blocks are the same height, then the Ui of the heavier block (on top) is 3 * g * z, where z is the distance between its initial height and the height where the two blocks meet. This makes the Ui of the lighter block 2 * g * (z-5), giving it negative potential energy. When the weight on the second block is removed, the lighter block will go up, and the heavier will go down. The U of both blocks is converted into kinetic energy.

I assumed that the initial energy was

3 * g * z + 2 * g * (z-5)

and the final was

K = 1/2 * mv^2

Both blocks have the same speed, so m = 3 + 2. This would describe the speed of the system at all points after U is converted completely into K. Does this happen when they are at the same height?

If this is the case, how would I find z?

 

[ashishsinghal]

yeah, your equations are correct:
The length of string is constant. Use this to obtain z.
Do you know that the speed is equal or you "assumed" it??

 

[DrWillVKN]

yeah, your equations are correct:
The length of string is constant. Use this to obtain z.
Do you know that the speed is equal or you "assumed" it??

I assumed it, because it only asks for 1 speed.

So, since they both go up to become 'equal', would z be equal to 5 - z, making z = 2.5 ?

EDIT: alright, it's the right answer, thanks!

 

[ashishsinghal]

I assumed it

Do you want its proof?