Question

A mass of 10 kg is dropped from a height of 50 cm . find its

potential energy just before dropping .

kinetic energy just on touching the ground .

velocity with which it hits the ground .


{given G=10m/s]

Answer 1

P.E before dropping = m x g x h
10 x 10 x 50 = 5000

K.E on touching the ground = 0
because, k.e = 1/2 x m x v²
on touching ground, v = 0
So, 1/2 x 10 x 0 = 0
,
Vel = 10 m/s

Answer 2

Explanation:

Given, mass of the object, m = 10 kg , Height, h = 50 cm= 0.5 m

(i) As potential energy is given by PE = mgh

= 10 x 10 X 0.5 = 50 J

(ii) From law of conservation of energy,

total energy of the ball just before dropping = total energy of the ball just on touching the ground => KE + PE of ball just before dropping

= KE of ball just on touching the ground = KE= 50 J

(iii)As , we know KE = 50J

so, 1/2 m {{v}^{2}} = 50

{{v}^{2}} = 50 x 2/10 = 10