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 |
| \[\Rightarrow \] \[W={{K}_{t}}-{{K}_{i}}\] |
| \[\Rightarrow \] \[{{K}_{f}}=W+{{K}_{i}}=\int_{{{x}_{1}}}^{{{x}_{2}}}{Fxdx}+\frac{1}{2}m{{v}^{2}}\] |
| \[=\int_{20}^{30}{-\,0.1\times \,dx}+\frac{1}{2}\times 10\times {{10}^{2}}\] |
| \[=-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{20}^{30}+500\] |
| \[=-0.05[{{30}^{2}}-{{20}^{2}}]+500\] |
| \[=-\,0.05\,[900-400]+500\] |
| \[\Rightarrow \] \[{{K}_{f}}=-\,25+500=475J\] |
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