A force acts for $0.1$ s 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 The force is then withdrawn and the body moves with a velocity of $2$ m/s. Find the magnitude of force.

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Solution

Given $: u = 0, t = 0.1 s, m = 2 k g, f i n a l v e l o c i t y v = 2 m / s; F = ?$
We know that acceleration is the rate of change of velocity
Acceleration $= \frac{(v - u)}{t}$
$= \frac{(2 - 0)}{0.1}$
$= 20 m / s^{2}$
$F o r c e = m a$
$= 2 \times 20$
$= 40 N$

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 The force is then withdrawn and the body moves with a velocity of 2m/s. Find the magnitude of force.

Given:u=0,t=0.1s,m=2kg,finalvelocityv=2m/s;F=?We know that acceleration is the rate of change of velocityAcceleration =(v−u)t=(2−0)0.1=20m/s2Force=ma=2×20=40N

A force acts for $0.1$ s 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 The force is then withdrawn and the body moves with a velocity of $2$ m/s. Find the magnitude of force.

Given $: u = 0, t = 0.1 s, m = 2 k g, f i n a l v e l o c i t y v = 2 m / s; F = ?$
We know that acceleration is the rate of change of velocity
Acceleration $= \frac{(v - u)}{t}$
$= \frac{(2 - 0)}{0.1}$
$= 20 m / s^{2}$
$F o r c e = m a$
$= 2 \times 20$
$= 40 N$