Question

A body of mass 5 kg under the action of constant force $\overrightarrow{F}=F_x\hat{i}+F_y\hat{j}$ has velocity at $t=0$s as $\overrightarrow{v}=(6\hat{i}-2\hat{j})m/s$ and at $t=10$s as $\overrightarrow{v}=6\hat{j}m/s$. The force $\overrightarrow{F}$ is :

• A. $(3\hat{i}-4\hat{j})N$
• B. $(-3\hat{i}+4\hat{j})N$
• C. $(\frac{3}{5}\hat{j}-\frac{4}{5}\hat{j})N$
• D. $(-\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j})N$

Answer 1

Answer: (B)

$(-3\hat{i}+4\hat{j})N$

We work this solution along the 2 axes.
For x - axis
$u=6$, $t=10sec$, $v=0$ If a be the acceleration of the particle.
Using Newton's equation of motion, $v=u+at$
$0=6+10a$ Thus, $a$ = $\frac{-3}{5}$
Similarly along the y axis
$u=-2$, $v=6$, $t=10$. If$A$ be the acceleration
$6=-2+10A,A=0.8$
Now, Using, Newton's second law,
$F=ma$
$\overrightarrow{F}=m\overrightarrow{a}$
$\overrightarrow{F}=5\times (-0.6i+0.8j)$
$\overrightarrow{F}=-3i+4j$