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

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(D) Let the acceleration of the container be $a_c = F/m$. In the frame of the container,the block experiences a pseudo-force $F_p = ma_c = F$ acting t

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the acceleration of the container be $a_c = F/m$. In the frame of the container,the block experiences a pseudo-force $F_p = ma_c = F$ acting towards the left.This is equivalent to a spring-mass system starting from its natural length with a constant force $F$ applied to the mass $m$. The effective spring constant is $k$.The angular frequency is $\omega = \sqrt{k/m}$.The time taken to reach maximum compression is half the time period of the oscillation,$t = T/2 = \pi / \omega = \pi \sqrt{m/k}$.During this time,the impulse equation for the system (container + block) in the horizontal direction is $\int F dt = \Delta P = (m+m)v_f - (m+m)v_i$.Since $v_i = 0$,we have $F \cdot t = 2mv_f$.Substituting $t = \pi \sqrt{m/k}$,we get $F \pi \sqrt{m/k} = 2mv_f$.Thus,$v_f = \frac{F \pi}{2} \sqrt{\frac{m}{k}} \cdot \frac{1}{m} = \frac{\pi F}{2} \sqrt{\frac{1}{km}}$.

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