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A chain is held on a frictionless table with one-fourth of its length hanging over the edge, as shown in Fig. 12-28. If the chain has a length L and a mass m, how much work is required to pull the hanging part back on the table?
A chain is held on a frictionless table with one-fourth of its length hanging over the edge, as shown in Fig. 12-28. If the chain has a length L and a mass m, how much work is required to pull the hanging part back on the table?

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5 years ago

Navjyot Kalra
654 Points
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5 years ago
Nitin
19 Points
```							Length of chain hanging from the table = L/4Mass of chain (only the hanging part)  = m/4Gravitation force on hanging part of the chain = mg/4Center of mass of the hanging part of the chain is at L/8 from end.( i.e, half of hanging lenght)therefore, Work done = F.s                               = mg/4 * L/8 cos180(degrees)                               =-mgL/32 Hence, external work done required to pull the hanging length of the chain =1/32mgL
```
3 years ago
618 Points
```							Dear student,Please find the attached solution to your problem. Let the surface of table be the datum, i.e. gravitational potential energy on table surface = 0The mass of hanging part = m/L * L/4 = m/4The Position of CM of hanging part = (L/4)/2 = L/8 from the surface of tableHence, Gravitational PE of the whole chain while ¼ th part of it is hanging = 0 + (m/4)g(-L/8) = – mgL/32When the chain is completely on the table, Gravitational PE = 0Now, work is done against gravity, which a conservative forcehence, W =  ΔU               =  (0 – ( – mgL/32))               = mgL/32 Hope it helpsThanks and regards,Kushagra
```
6 months ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions