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Sidhant Thakur , 5 Years ago
Grade 11
Mechanics> HelpHelpHelpHelpHelpHelpHelpHelpHelpHelpH...
2 Answers
Arun
Last Activity: 5 Years ago
The work done to lift the hanging part of the cable up to the surface is MgL/2n²
Given-
Mass of uniform cable = M
Length of cable = L
Length of cable hanging below the edge of the surface is = 1/n th part
Mass of the hanging part = M/n
Center of mass of h = L/2n
So work done should be-
W = mgh = (M/n) g (L/2n)
W = MgL/2n²
Given-
Mass of uniform cable = M
Length of cable = L
Length of cable hanging below the edge of the surface is = 1/n th part
Mass of the hanging part = M/n
Center of mass of h = L/2n
So work done should be-
W = mgh = (M/n) g (L/2n)
W = MgL/2n²
Vikas TU
Last Activity: 5 Years ago
Dear student
There is short cut.
Work done against gravity = mgh
= m/n g l/2n
= mgl /2n^2
Work done against gravity = mgh
= m/n g l/2n
= mgl /2n^2
Good Luck
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