Question

A $0.2$ kg object is subjected to a force of $3\overrightarrow{i}-0.4\overrightarrow{j}$ newton. What is velocity vector after $6$ secs?

Answer 1

We know Force F=mass$\times$acceleration

and acceleration a$=\frac{\text{Final velocity(v)- initial velocity(u)}}{\text{time(t)}}$

$F=\text{m}\frac{\text{v}}{\text{t}}\overrightarrow{v}=\frac{\overrightarrow{\text{F}}\times \text{t}}{\text{m}}=\frac{6\times (3\hat{i}-0.4\hat{j})}{0.2}=90\hat{i}-12\hat{j}$