A) 475 J
B) 450 J
C) 275 J
D) 250 J
Correct Answer: A
Solution :
From work-energy theorem, Work done = Change in KE \[\Rightarrow \,\,\,W={{K}_{t}}-{{K}_{i}}\] \[\Rightarrow \,\,\,{{K}_{f}}=W+{{K}_{i}}=\int_{{{x}_{1}}}^{{{x}_{2}}}{Fxdx}+\frac{1}{2}m{{v}^{2}}\] \[=\int_{20}^{30}{-0.1x\,dx}+\frac{1}{2}\times 10\times {{10}^{2}}\] \[=-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{20}^{30}+500\] \[=-0.05[{{30}^{2}}-{{20}^{2}}]+500\] \[=-0.05[900-400]+500\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{f}}=-25+500=475J\]
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