A uniform cable of mass M and length L is placed on a horizontal surface such that its 1 n t h part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

Mass of the part hanging is M n . Centre of mass of this part lies L 2 n distance from the surface. ∴ Work done in lifting the hanging part to surface is W = M n · g L 2 n = M g L 2 n 2

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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Calculation of workHard2 minQB

PhysicsHardsingle choice

A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its ${(\frac{1}{n})}^{t h}$ part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

Options:

Answer:

Solution:

Mass of the part hanging is $\frac{M}{n}$ . Centre of mass of this part lies $\frac{L}{2 n}$ distance from the surface.
solution
$∴$ Work done in lifting the hanging part to surface is
$W = (\frac{M}{n}) \cdot g (\frac{L}{2 n})$
$= \frac{M g L}{2 n^{2}}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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