Question

A 50kg boy runs up the steps to the third floor of height 10.5m in 45s.While at the third floor,the boy drpos a 0.25kg ball and strike the ground.
1)calculate the rate of work done
2)by using the work-energy theorem,calculate the speed of the ball just before it strikes the ground
3)calculate the average force applied on the ball if the impulsive force has acted for 4ms.(assume no energy loss)

Answer 1

1. F = M*g = 50 * 9.8 = 490 N. = Wt. of boy.
P = F*d/t = 490 * 10.5/45 = 114.3 J/s = 114.3 Watts. = Work rate.

2. KE = PE.
0.5M*V^2 = M*g*h
0.50*0.25*V^2 = 0.25**9.8*10.5
V = ?.

Answer 2

In the third solution:
F=mv/t
m=0.25 kg and t=4ms=0.004 s.
\\
correct.