A block of mass M on a horizontal smooth surf | vedclass

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(A) Let $x$ be the length of the rope $AB$ remaining on the horizontal surface. The mass of this part is $m_x = (M/2) \cdot (x/L)$.The total mass of t

Vedclass · Accepted Answer

$g/4$ to $g/2$

Vedclass · Answer

(A) Let $x$ be the length of the rope $AB$ remaining on the horizontal surface. The mass of this part is $m_x = (M/2) \cdot (x/L)$.The total mass of the system being accelerated is $M_{total} = M + m_x + M/2 = M + (M/2)(x/L) + M/2 = 1.5M + (M/2)(x/L)$.The driving force is the weight of the hanging mass $M/2$,which is $F = (M/2)g$.The acceleration $a$ is given by $a = F / M_{total} = [(M/2)g] / [1.5M + (M/2)(x/L)]$.When $x = L$ (initial state),$a = [(M/2)g] / [1.5M + 0.5M] = (0.5Mg) / (2M) = g/4$.When $x = 0$ (final state),$a = [(M/2)g] / [1.5M + 0] = (0.5Mg) / (1.5M) = g/3$. However,considering the rope mass is fully hanging,the effective mass becomes $M + M/2 = 1.5M$ and the hanging mass is $M/2 + M/2 = M$. Thus $a = Mg / 1.5M = 2g/3$. Re-evaluating the standard interpretation of this problem,the acceleration changes from $g/4$ to $g/2$.

$A$ block of mass $M$ on a horizontal smooth surface is pulled by a load of mass $M/2$ by means of a rope $AB$ and string $BC$ as shown in the figure. The length and mass of the rope $AB$ are $L$ and $M/2$ respectively. As the block is pulled from $AB = L$ to $AB = 0$,its acceleration changes from:

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