Question

A body of mass 3 kg is dropped from a height of 1m. The K.E of the body before touching the ground is:

• A. 150 J
• B. 30 N
• C. 30 J
• D. 150 N

Answer 1

The correct option is C

30 J

Given:
Mass of the body m = 3 kg
Initial height of the body from the ground, h = 1 m
Since the body is dropped, its initial velocity u = 0
According to law of conservation of energy

Initial K.E + Initial P.E = Final K.E + Final P.E

or, 0 + $m\times g\times \left(1m\right)$ = Final K.E +0

or, Final K.E =$3\times 10\times \left(1m\right)$ = 30 joules