A) 475 J
B) 450 J
C) 275 J
D) 250 J
Correct Answer: A
Solution :
Apply work-energy theorem, When a force acts upon a moving body, then the kinetic energy of the body increases and the increase is equal to the work done. This is work-energy theorem. \[Work\,\,done\,\,=\frac{1}{2}m{{\nu }^{2}}-\frac{1}{2}m{{u}^{2}}={{K}_{f}}-{{K}_{i}}\] Another definition of work done is force x displacement. \[\therefore \,\,\,\,\,\,\,Fdx={{K}_{f}}-\frac{1}{2}m\nu _{i}^{2}\] where the subscripts f and i stand for final and initial. \[F.dx\,\,=\,\,{{K}_{f}}-\frac{1}{2}\times 10\times {{(10)}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\operatorname{F}.dx={{K}_{f}}\,-\,\,500\] \[\Rightarrow \,\,\,\,\,\,\,\,\int\limits_{x\,=\,20}^{x\,=\,30}{(0.1)x\,dx={{K}_{f}}-500}\] Using the formula \[\int{{{x}^{n}}dx}=\frac{{{x}^{n+1}}}{n+1}\] we have \[-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{x=20}^{x=30}={{K}_{f}}-500\] \[-0.1\left[ \frac{{{(30)}^{2}}}{2}-\frac{{{(20)}^{2}}}{2} \right]\,\,={{K}_{f}}-500\] \[\Rightarrow \,\,\,\,\,\,{{\operatorname{K}}_{f}}-500=-\,25\] \[\Rightarrow \,\,\,\,\,\,{{\operatorname{K}}_{f}}=500 -25- 475 J\]
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