Pulley, conservation of energy

[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?

 

[rwisz]

Your approach to solving this problem using the conservation of energy principle is correct. However, there are a few things that could be clarified.

Firstly, when determining the initial potential energy of the lighter block, it should be noted that its initial height is 5m lower than the heavier block, so the potential energy should be calculated as 2 * g * (z-5).

Secondly, to find z, you can use the fact that at the point where the two blocks are at the same height, their potential energies will be equal. So, you can set the initial potential energy of the lighter block equal to the final potential energy of the heavier block, and solve for z.

Finally, to determine the speed of the blocks when they are at the same height, you can use the conservation of energy equation again. Set the initial potential energy of the system (3 * g * z + 2 * g * (z-5)) equal to the final kinetic energy of the system (1/2 * (3+2) * v^2) and solve for v. This will give you the speed of the blocks when they are at the same height.

Overall, your approach is correct, but be sure to consider the initial potential energy of the lighter block and use the conservation of energy equation to solve for both z and the final speed of the blocks.