A) 0.5 J
B) 1.5 J
C) 2.0 J
D) 2.5 J
Correct Answer: A
Solution :
[a] : Here. \[m=1\text{ }kg,{{v}_{i}}=2\text{ }m\text{ }{{\text{s}}^{-1}},k=0.5\text{ }J\]Initial kinetic energy, \[{{K}_{i}}=\frac{1}{2}mv_{i}^{2}=\frac{1}{2}\times (1kg){{(2m{{s}^{-1}})}^{2}}=2J\] Work done by retarding force \[W=\int_{{}}^{{}}{{{F}_{r}}dx}=\int\limits_{0.1}^{2.01}{-\frac{k}{x}}dx=-k\left[ \ln \,x \right]_{0.1}^{2.01}\] \[=-k\ln \left( \frac{2.01}{0.1} \right)=-0.5\ln (20.1)=-1.5J\] According to work-energy theorem\[W={{K}_{f}}-{{K}_{i}}\] or \[{{K}_{f}}=W+{{K}_{i}}=-1.5J+2J=0.5J\]
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