What is the maximum height reached by the 2kg mass in a modified Atwood machine?

[serverxeon]

Homework Statement

Two masses, 5kg and 2kg on the two ends of an atwood machine (ie. over a pulley).
Initially, both are 0.6m above ground.

Find the maximum height the 2kg mass will reach when the system is released.

Homework Equations

The Attempt at a Solution

Maximum height will = initial 0.6 m + 0.6m it will move up + the "shooting up effect" due to KE it still has at h=1.2m

If i use net force to get net acceleration, and thereby get the velocity when 0.6m has been moved, i will arrive at an additional displace of 0.26 m

Correct answer is 1.46m, and so I'm right.

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But , if I use energy methods, I am stuck.
Total GPE lost by 5kg block = 29.4J
So, that means the 2kg block will gain a height of 29.4/(2*9.8) = 1.5m?

I think I am missing out something. Can the energy be 'transferred' from m1 to m2 in this case? Or is energy lost to heat when the 5kg smashes the floor?

Anyone has a way to solve this using energy methods?

 

[gneill]

When the 2kg block has risen 0.6m and the 5kg block has fallen 0.6m you can determine the net KE gained by the system from the difference in PE changes.

KE = (5kg - 2kg)*g*0.6m

This turns out to be about 17.65J.

Now, this energy is divided between the two masses. The 5kg block, being more massive, has the larger share. The 2kg block has the smaller share. In particular,

KEs = KE*2/(5 + 2) = 5.04J

The energy that goes with the larger block is "lost" to us when it hits the ground. The energy in the smaller block carries on, raising it above the "launch" point.