Question

A 40 newton object is released from a height of 10 m. Just before it hits the ground, its kinetic energy, in joules is

Answer 1

Dear student
Given - force = 40N
height = 10m
solution : from third law of motion
$v^2-u^2=2gh$
u=0 ; $v^2=2\times 10\times 10=200m/s$
as F= mg
40 = m$\times$10
m = 4kg
K.E. =$\frac{1}{2}m{v}^2$ = $\frac{1}{2}\times 4\times 200$ = 400joules
Regards