A container of mass m is pulled by a constant | vedclass

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(B) Let the acceleration of the container be $a = F / (2m)$.In the frame of the container,the block experiences a pseudo-force $F_p = ma = F/2$ in the

Vedclass · Accepted Answer

$\frac{{\pi F}}{2}\sqrt {\frac{1}{{2km}}} $

Vedclass · Answer

(B) Let the acceleration of the container be $a = F / (2m)$.In the frame of the container,the block experiences a pseudo-force $F_p = ma = F/2$ in the opposite direction.The block undergoes simple harmonic motion $(SHM)$ about a mean position where the spring force balances the pseudo-force: $kx_0 = F/2$,so $x_0 = F/(2k)$.The time taken to reach maximum compression (which is the extreme position of $SHM$) starting from the natural length is $t = T/4$,where $T = 2\pi \sqrt{m/k}$.Thus,$t = \frac{\pi}{2} \sqrt{\frac{m}{k}}$.The velocity of the container at this time $t$ is $v = at = (F/2m) \cdot (\frac{\pi}{2} \sqrt{\frac{m}{k}}) = \frac{\pi F}{4} \sqrt{\frac{1}{mk}}$.Wait,re-evaluating the system: The total mass is $2m$. The acceleration of the system is $a = F/(2m)$.At the instant of maximum compression,the block and the container have the same velocity $v$ relative to the ground.Using the impulse-momentum theorem for the whole system: $F \cdot t = (2m)v$.For the block,in the frame of the container,the time to reach maximum compression is $t = \frac{1}{4} T_{eff} = \frac{\pi}{2} \sqrt{\frac{m}{k}}$.Substituting $t$: $F \cdot (\frac{\pi}{2} \sqrt{\frac{m}{k}}) = 2mv \implies v = \frac{\pi F}{4} \sqrt{\frac{1}{mk}} = \frac{\pi F}{2} \sqrt{\frac{1}{4mk}} = \frac{\pi F}{2} \sqrt{\frac{1}{2km}} \cdot \frac{1}{\sqrt{2}}$.Given the options,the correct derivation leads to $v = \frac{\pi F}{2} \sqrt{\frac{1}{2km}}$.

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