A uniform rope of length 4 ,m and mass 0.4 ,k | vedclass

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(D) The mass per unit length of the rope is $\lambda = \frac{M}{L} = \frac{0.4 \,kg}{4 \,m} = 0.1 \,kg/m$.The length of the hanging part is $l = 0.6 \

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$0.18$

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(D) The mass per unit length of the rope is $\lambda = \frac{M}{L} = \frac{0.4 \,kg}{4 \,m} = 0.1 \,kg/m$.The length of the hanging part is $l = 0.6 \,m$.The mass of the hanging part is $m = \lambda 	imes l = 0.1 \,kg/m 	imes 0.6 \,m = 0.06 \,kg$.The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6 \,m}{2} = 0.3 \,m$ below the table edge.The work done to pull the rope onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.Substituting the values: $W = 0.06 \,kg 	imes 10 \,m/s^2 	imes 0.3 \,m = 0.18 \,J$.

$A$ uniform rope of length $4 \,m$ and mass $0.4 \,kg$ is held on a frictionless table in such a way that $0.6 \,m$ of the rope is hanging over the edge. The work done to pull the hanging part of the rope onto the table is,(Assume $g = 10 \,m/s^2$) (in $\,J$)

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