6.A uniform chain of length 2m is kept on a table such that length of 60cm hangs freely from the edge of a table.The total mass of the chain is 4kg.What is the work in pulling the entire chain on the table ?-------
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Given
M and L are total lenght and mass of the chain.
M = 4 kg
L = 2 m
And let length and mass of part of chain hanging down are l and m respectively
l = 0.6 m
considering mass per unit length =>
M/L = m/l
m = l / L * 4 = (0.6 / 2) * 4 = 1.2 kg
Tension in chain hanging down = mg = 11.76 N (g=9.8 m/s^2)
Force required = Tension
Displacement = 0.6 m (part of chain hanging comes up)
W= Force * Displacement = 11.76 * 0.6 = 7.056 J
Hope it helps!!!
M and L are total lenght and mass of the chain.
M = 4 kg
L = 2 m
And let length and mass of part of chain hanging down are l and m respectively
l = 0.6 m
considering mass per unit length =>
M/L = m/l
m = l / L * 4 = (0.6 / 2) * 4 = 1.2 kg
Tension in chain hanging down = mg = 11.76 N (g=9.8 m/s^2)
Force required = Tension
Displacement = 0.6 m (part of chain hanging comes up)
W= Force * Displacement = 11.76 * 0.6 = 7.056 J
Hope it helps!!!
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