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A uniform rope of length \(4 \mathrm{~m}\) and mass \(0.4 \mathrm{~kg}\) is held on a frictionless table in such a way that \(0.6 \mathrm{~m}\) of the rope is hanging over the edge. The work done to pull the hanging part of the rope on the to the table is, (Assume \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )
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Verified Answer
The correct answer is:
\(0.18 \mathrm{~J}\)
\(\begin{aligned}
& \text {Hint : } W=\frac{\mathrm{mgL}}{2}=\frac{0.4}{4} \times 0.6 \times 10 \times \frac{0.6}{2} \\
& =0.1 \times 0.6 \times 10 \times 0.3 \\
& =0.18 \mathrm{~J}
\end{aligned}\)
& \text {Hint : } W=\frac{\mathrm{mgL}}{2}=\frac{0.4}{4} \times 0.6 \times 10 \times \frac{0.6}{2} \\
& =0.1 \times 0.6 \times 10 \times 0.3 \\
& =0.18 \mathrm{~J}
\end{aligned}\)
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