Solving Pushing Two Blocks Kinetic Energy

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SUMMARY

The discussion centers on calculating the final kinetic energy of two blocks, A and B, with masses of 4 kg and 1 kg, respectively, when a force pushes block A. The initial speed of both blocks is 2.0 m/s, and the work done on block A is 10 J. The correct final kinetic energy for block A is determined to be 18 J, while for block B, it is 12 J. A key insight is that both blocks move together, necessitating an analysis of the system's total kinetic energy change before determining individual energies.

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Homework Statement



There are two blocks touching each other, block A, and block B. Block A is being pushed by a force, which in turn moves block A and B at the same time. Block A has a mass of 4kg, block B has a mass of 1kg. (so the force is pushing the larger block, which is pushing the smaller block).

When the hand starts to push, the blocks are moving with a speed of 2.0 m/s. Suppose that the work done by the hand on block A during a given displacement is 10 J. Determine the final kinetic energy of each block



Homework Equations



Change in work (f * d) = change in kinetic energy ( 1/2m(vf)^2 - 1/2m(vi)^2)


The Attempt at a Solution



10J = Kf - 1/2(4kg)(2m/s)^2, Kf = 18J

10J = Kf - 1/2(1kg)(2m/s)^2, Kf = 12J

Now to me, this seems like a pretty straight forward problem. However, my fellow students are getting a different answer. Perhaps I am overlooking something? Any help is appreciated.
Thanks
 
Physics news on Phys.org
Since the blocks move together at the same speeds, you've got to first look at the change in KE of the system (both blocks together) before you calculate the individual final KE of each.
 

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