## Problem Statement

**A force acts for 0.1 sec on a body of mass 2 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m/s. Find the magnitude of the force.**

## Solution of class 9 Set1 Q12

** Solution**:

A force acts on a body for 0.1 sec. That means that body had acceleration for 0.1 sec.

So time duration = t = 0.1 sec

Initial velocity =u= 0

When the force is withdrawn, the velocity
becomes 2 m/s.

That means, at the end of accelerated motion the final velocity = v=2 m/s

(Now, to find out the magnitude of force, we need to find out the value of acceleration. mass is already given)

Acceleration a = (v-u)/t = (2-0)/0.1= 20 m/s^2

And the magnitude of force (F) = ma= mass.Acceleration = 2 X 20 N = 40 N

Answer: Magnitude of Force = 40 Newton