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Grade: upto college level

                        

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?

5 years ago

Answers : (3)

Navjyot Kalra
askIITians Faculty
654 Points
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232-1830_1.JPG
5 years ago
Nitin
19 Points
							
Length of chain hanging from the table = L/4
Mass of chain (only the hanging part)  = m/4
Gravitation force on hanging part of the chain = mg/4
Center 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
Kushagra Madhukar
askIITians Faculty
605 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 = 0
The mass of hanging part = m/L * L/4 = m/4
The Position of CM of hanging part = (L/4)/2 = L/8 from the surface of table
Hence, Gravitational PE of the whole chain while ¼ th part of it is hanging = 0 + (m/4)g(-L/8) = – mgL/32
When the chain is completely on the table, Gravitational PE = 0
Now, work is done against gravity, which a conservative force
hence, W =  ΔU
               =  (0 – ( – mgL/32))
               = mgL/32
 
Hope it helps
Thanks and regards,
Kushagra
3 months ago
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