First slide

Work done by Diff type of forces

Question

A uniform chain of length 2m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. The work done in pulling the entire chain on the table is (Take g = ${ms}^{- 2}$ )

Moderate

A

12.9 J
B

6.3 J
C

3.6 J
D

2.0 J

Solution

Mass of length 2 m of the chain = 4 kg
Mass of length 60 cm or 0.60 m of the chain = $\frac{4 \times 0.60}{2} = 1.2 kg$
Weight of the hanging part of the chain = 1.2 $\times$ 10 = 12 N
Since of centre of gravity of the hanging part lies at its mid-point, i.e., 30 cm or 0.30 m
W = $12 \times 0.30 = 3.6 J$

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