A force acts for 0.1 s on a body of mass 2.0 Kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m s^-1. Find the magnitude of force.

The 1^st equation of motion states that;

v = u + at

Given,

t = 0.1 s

m = 2.0 Kg

u = 0

v = 2 m s^-1

Substituting the values in the formula above we get,

$2 = 0 + (a \times 0.1) \\[0.5em] \Rightarrow a = \dfrac{2}{0.1} \\[0.5em] \Rightarrow a = 20 \text {m s}^{-2} \\[0.5em]$

Hence, acceleration of the body = 20 m s^-2`

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$F = 2 \times 20 \\[0.5em] \Rightarrow F = 40 N \\[0.5em]$

Hence,

Magnitude of force = 40 N