A uniform cable of mass M and length L is pla | vedclass

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(B) The length of the hanging part is $l = L/n$.Since the cable is uniform,the mass of the hanging part is $m = (M/L) 	imes l = M/n$.The center of ma

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$\frac{MgL}{2n^2}$

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(B) The length of the hanging part is $l = L/n$.Since the cable is uniform,the mass of the hanging part is $m = (M/L) 	imes l = M/n$.The center of mass of the hanging part is at a distance $h_{COM} = l/2 = L/(2n)$ below the edge of the surface.To lift the hanging part onto the surface,we must raise its center of mass to the level of the surface.The work done $W$ is equal to the change in potential energy of the hanging part:$W = m 	imes g 	imes h_{COM}$$W = (M/n) 	imes g 	imes (L/(2n))$$W = \frac{MgL}{2n^2}$

$A$ uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface,the work done should be

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