A particle of mass m is at rest in a train mo | vedclass

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(D) Work done is defined as $W = \int \vec{F} \cdot d\vec{s}$.Let the train move with velocity $v$ with respect to the ground.In the train's frame,the

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

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(D) Work done is defined as $W = \int \vec{F} \cdot d\vec{s}$.Let the train move with velocity $v$ with respect to the ground.In the train's frame,the initial velocity of the particle is $0$. The acceleration is $a = F_0/m$. The displacement in time $t_0$ is $x = \frac{1}{2} a t_0^2 = \frac{F_0 t_0^2}{2m}$.The work done measured by the girl in the train is $W_g = F_0 x = \frac{F_0^2 t_0^2}{2m}$.In the ground frame,the initial velocity of the particle is $v$. The displacement in time $t_0$ is $x' = v t_0 + \frac{1}{2} a t_0^2 = v t_0 + x$.The work done measured by the boy on the ground is $W_b = F_0 x' = F_0(v t_0 + x) = F_0 v t_0 + W_g$.Since $F_0, v, t_0 > 0$,it follows that $W_b > W_g$.Therefore,the statements that 'Both will measure the same work' and 'Girl will measure higher value than the boy' are incorrect.Thus,the incorrect options are $(A)$ and $(C)$.

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