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 2.0 kg so that it is moving with $4m/s^2$ in east direction. If $\overrightarrow{F_1}$ force is removed, then the acceleration becomes 7 $m/s^2$ in north, then the acceleration of the block if only $\overrightarrow{F_1}$ is acting will be :

• A. $16m/s^2$
• B. $\sqrt{65}m{s}^2$
• C. $\sqrt{260}m{s}^2$
• D. $\sqrt{3}3m{s}^2$

Answer 1

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

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=ma$

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=4\times 2=8i........\left(1\right)$

Now, when force $F_1$ is removed then,

$\overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=ma$

$\overrightarrow{F_2}+\overrightarrow{F_3}+\overrightarrow{F_4}+\overrightarrow{F_5}=7\times 2=14j..........\left(2\right)$

$\left(i\right)$ & $\left(ii\right)$

$F_1=8i-14j$

$a_1=\frac{F_1}{m}$

$=\frac{8i-14j}{2}=4i-4j$

$=\sqrt{65}$

$\therefore$ Option $B$ is correct.