Question

A pushing force of 12N is applied on a box for 32ms. What is the magnitude of the impulse of this force?

Answer 1

Given data:

Force;

$F=12\text{ N}$

Time;

$\begin{gathered} T=32\text{ ms} \\ =32\times10^{-3}\text{ s} \end{gathered}$

The impulse is given as,

$\Delta p=F\times T$

Substituting all known values,

$\begin{gathered} \Delta p=(12\text{ N})\times(32\times10^{-3}\text{ s}) \\ =0.384\text{ N}\cdot\text{ s} \end{gathered}$

Therefore, the magnitude of the impulse of this force is 0.384 N.s.