Question

A body of mass 10 kg falls from a height of 5m and stopped within one-tenth of a second on the ground .the force of interaction is

Answer 1

Initial velocity of the body = 0 m/s

Height through which it falls = 5 m

Acceleration due to gravity = 9.8 m/s^2

Let the velocity of the body at the bottom be v.

Using energy conservation:

$\frac{1}{2} \times m \times v^2 = mgh$

$v = \sqrt{2 \times 9.8 \times 5}$

v = 9.9 m/s

Now, the force of interaction = mg + Rate of change of momentum

Rate of change of momentum = $\frac{10 \times 9.9}{0.1}$

Rate of change of momentum = 990 N

Force of interaction = 990 + 98

Force of interaction = 1088 Newtons