Question

Five forces ${\overrightarrow{F}}_1$, ${\overrightarrow{F}}_2$, ${\overrightarrow{F}}_3$, ${\overrightarrow{F}}_4$ and ${\overrightarrow{F}}_5$ are acting on a particle of mass $2.0kg$ so that it is moving with $4m/s^2$ in east direction. If $\overrightarrow{F_1}$ force is removed, then the acceleration becomes $7m/s^2$ in north, then the acceleration of the block if only $\overrightarrow{F_1}$ is acting will be:

• A. $16m/s^2$
• B. $\sqrt{33}m/s^2$
• C. $\sqrt{65}m/s^2$
• D. $\sqrt{260}m/s^2$

Answer 1

The correct option is C $\sqrt{65}m/s^2$

Given, mass of the particle, $m=2kg$
And acceleration under forces $\overrightarrow{F_1},\overrightarrow{F_2},\overrightarrow{F_3},\overrightarrow{F_4}$ and $\overrightarrow{F_5}$ is,
$\overrightarrow{a}=4\hat{i}$

Now we know,
$\overrightarrow{F}=m\overrightarrow{a}$
$\Rightarrow \overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=2\left(4\hat{i}\right)\cdot \cdot \cdot \left(i\right)$

If force $\overrightarrow{F_1}$ is removed, then acceleration, $\overrightarrow{a}=7\hat{j}$
$\Rightarrow \overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=2\left(7\hat{j}\right)\cdot \cdot \left(ii\right)$

From (i) and (ii),
$\overrightarrow{F_1}=8\hat{i}-14\hat{j}$

So, acceleration due to force $\overrightarrow{F_1}$,
$\overrightarrow{a_1}=\frac{\overrightarrow{F_1}}{m}=4\hat{i}-7\hat{j}$

$\Rightarrow a_1=\sqrt{16+49}$

$a_1=\sqrt{65}m/s^2$

Final answer: (b)