A) 7.2 J
B) 3.6 J
C) 120 J
D) 1200 J
Correct Answer: B
Solution :
[b] Mass per unit length\[=\frac{M}{L}=\frac{4}{2}=2\text{ }kg/m\] |
The mass of 0.6 m of chain\[=0.6\times 2=1.2\text{ }kg\] |
The centre of mass of hanging part |
\[h=\frac{0.6+0}{2}=0.3\,m\] |
Hence, work done in pulling the chain on the table = work done against gravity force |
\[W=mgh=1.2\times 10\times 0.3=3.6J\] |
Alternative Method |
As\[\frac{60\,cm}{200\,cm}=\frac{3}{10}\] part of chain is hanging from the table, the change in potential energy in pulling hanging part onto the table. |
\[\Delta {{U}_{1}}=-\int\limits_{0}^{\frac{3}{10}l}{\frac{m}{l}g}\,xdx\]\[(\because \Delta U=W=F.dx=mg.dx)\]\[=\frac{m}{l}\,g\,\int_{0}^{\frac{3}{10}l}{xdx=\frac{m}{l}g\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{\frac{3}{10}l}=\frac{m}{l}\frac{g}{2}{{\left( \frac{3}{10}l \right)}^{2}}}\] |
\[=-\frac{9mgl}{200}=-3.6J\] |
Now, work done\[=-\Delta U=-(-3.6J)\] |
\[=3.6\,J\] |
(suppose potential energy level at table surface is zero) |
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