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.

net work = change in kinetic energy

for Block B, we just have the force from block A acting on it

F (ab) d =.5 (1) vf² -.5 (1) (2²)

F (ab) d =.5vf² - 2

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction

10-F (ba) d =.5 (4) vf² -.5 (4) (2²)

10-F (ba) d = 2vf² - 8

F (ba) d = 18 - 2vf²

now we have two equations:

F (ba) d = 18 - 2vf²

F (ab) d =.5vf² - 2

since the magnitude of F (ba) and F (ab) is the same, substitute and find vf (I already took into account the direction when solving for F (ab)

10-.5vf² + 2 = 2vf² - 8

12 -.5vf² = 2vf² - 8

20 = 2.5vf²

vf² = 8

they both will have the same velocity

KE of block A =.5 (4) (2.828²) = 16 J

KE of block B=.5 (1) (2.828²) = 4 J