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Q. Five forces $\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3}, \vec{F}_{4}$ and $\vec{F}_{5}$ are acting on a particle of mass $2.0 \,kg$ so that it is moving with $4\,m/s^{2}$ in east direction. If $\vec{F}_{1}$ force is removed, then the acceleration becomes $7\,m/s^{2}$ 2 in north, then the acceleration of the block if only $\vec{F}_{1}$ is acting will be $\sqrt{n}\,m /s^{2}$ Find the value of n

Laws of Motion

Solution:

$\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\vec{F}_{4}+\vec{F}_{5} = 2\left(4i\right) \ldots\left(i\right)$
and $\vec{F}_{2}+\vec{F}_{3}+\vec{F}_{4}+\vec{F}_{5}=2\left(7\,j\right) \ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, $\vec{F}_{1}=8\hat{i}-14 \hat{j} $
$\vec{a}_{1}=\frac{\vec{F}_{1}}{m}=4\hat{i}-7 \hat{j} $
$\Rightarrow a_{1}=\sqrt{16+49}=\sqrt{65}\, m/ s^{2}$