Question

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

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

Answer 1

The correct option is A $(-3\hat{j}+4\hat{j})\text{N}$

From question,
Mass of body, $m=5\text{kg}$
Velocity at $t=0$ is $u=6\hat{i}-2\hat{j}\text{m/s}$
Velocity at $t=10\text{s}$ is $v=+6\hat{j}\text{m/s}$

Acceleration, $a=\frac{v-u}{t}$
$=\frac{6\hat{j}-(6\hat{i}-2\hat{j})}{10}=\frac{-3\hat{i}+4\hat{j}}{5}{\text{m/s}}^2$

Now, using, Newton's second law,

$F=ma$
$=5\times \frac{-3\hat{i}+4\hat{j}}{5}=(-3\hat{i}+4\hat{j})\text{N}$

Hence, $\left(A\right)$ is the correct answer.