Question

The force F is exerted on a particle of mass $5$ kg. If the particle starts from rest from origin and its position after $3$ s is $(4.5\hat{i}+13.5\hat{j})m$, then find $\overrightarrow{F}$.

• A. $(5\hat{i}+13.5\hat{j})N$
• B. $(5\hat{i}+15\hat{j})N$
• C. $(5\hat{i}+13\hat{j})N$
• D. $(15\hat{i}+5\hat{j})N$

Answer 1

The correct option is B $(5\hat{i}+15\hat{j})N$

Along $X$ direction displacement is $4.5m$ and in $Y$ it is $13.5m$.

Let the initial velocity be $u=0$.

So, $S=ut+a{t}^2/2$

Now, $4.5=a_x{3}^2/2$

This gives: $a_x=1m/s^2$

also, $13.5=a_y{3}^2/2$

$a_y=3m/s^2$

Now, $a=a_x+a_y=1\hat{i}+3\hat{j}$

$F=ma=$$5\hat{i}+15\hat{j}$, $m=5kg$