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*10^(-3)\text{ s} \end{gathered}`

The impulse is given as,

`\Delta p=F* T`

Substituting all known values,

`\begin{gathered} \Delta p=(12\text{ N})*(32*10^(-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.