Question

A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are moving with a speed of 2 m/s. Suppose that the work done on block A by the hand during a given displacement is 10 J. Determine the final energy of each block. Mass of A is 4kg, mass of B is 1kg.

Answer 1

It seems like you'd end up with \[1/2m _{A}v ^{2}+1/2m _{B}v ^{2}=0\] And that doesn't make much sense at all.

Answer 2

The initial velocity was 2m/s, so there are the terms:\[\frac{1}{2}M_{A}2^2 and \frac{1}{2}M_{B}2^2\]The work done on an object is equal to the variation of the kinectic energy, so the final energy is:
\[E=E_{initial}+W=2M_{A}+2M_{B}+W_{A}+W_{B}\]
Now to determina Wb, that is the only one we dont have, we need to find how much the velocity changed. To do that, we consider Wa and Ea.
\[\frac{1}{2}M_{A} ( v_{final}^2-v_{initial}^2)=W_{A}=10J\]From this we get: vf=3m/s
Since the initial velocity and the final velocity are the same for both objects W only depends on the mass, and since mass A is 4 times mass B, Wb=Wa/4=5/2
Now just put in the second formula:
\[E=2M_{A}+2M_{B}+10+\frac{5}{2}=8+2+10+\frac{5}{2}=22.5\]

Answer 3

Thanks so much! Now I understand.