weight = 50g = 490 N
friction force = .25 * 490 = 122.5 N
net force = 150 - 122.5 = 27.5 N
force * distance = 27.5*5 = 137.5 Joules work input
.5 m v^2 = 137.5
v^2 = 5.5
v =2.35 at the end
A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
A) What is the total work done on the block?
B) If the box started from rest, what is the final speed of the block?
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