Five persons A, B, C, D and E are pulling a c | vedclass

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(C) Let the total force exerted by the five persons be $\vec{F}_{total} = \vec{F}_A + \vec{F}_B + \vec{F}_C + \vec{F}_D + \vec{F}_E$. Given $m = 100 \

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Let the total force exerted by the five persons be $\vec{F}_{total} = \vec{F}_A + \vec{F}_B + \vec{F}_C + \vec{F}_D + \vec{F}_E$. Given $m = 100 \, kg$ and $\vec{a} = 3 \hat{i} \, m/s^2$ (taking East as $+\hat{i}$),we have $\vec{F}_{total} = 100 	imes 3 \hat{i} = 300 \hat{i} \, N$ $...(1)$.When $A$ stops,the force is $\vec{F}_{total} - \vec{F}_A = 100 	imes (-1 \hat{i}) = -100 \hat{i} \, N$ $...(2)$.Subtracting $(2)$ from $(1)$,we get $\vec{F}_A = 400 \hat{i} \, N$.When $B$ stops,the force is $\vec{F}_{total} - \vec{F}_B = 100 	imes (24 \hat{j}) = 2400 \hat{j} \, N$ (taking North as $+\hat{j}$) $...(3)$.Subtracting $(3)$ from $(1)$,we get $\vec{F}_B = 300 \hat{i} - 2400 \hat{j} \, N$.When only $A$ and $B$ pull,the net force is $\vec{F}_{net} = \vec{F}_A + \vec{F}_B = 400 \hat{i} + 300 \hat{i} - 2400 \hat{j} = 700 \hat{i} - 2400 \hat{j} \, N$.The acceleration is $\vec{a} = \frac{\vec{F}_{net}}{m} = \frac{700 \hat{i} - 2400 \hat{j}}{100} = 7 \hat{i} - 24 \hat{j} \, m/s^2$.The magnitude is $|\vec{a}| = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, m/s^2$.

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