A) 475 J
B) 450 J
C) 275 J
D) 250 J
Correct Answer: A
Solution :
The work, \[W=K{{E}_{F}}-{{K}_{{{E}_{I}}}}\] \[W=K{{E}_{F}}-\frac{1}{2}m{{v}^{2}}\] \[F.dx=K{{E}_{F}}-\frac{1}{2}\times 10\times {{10}^{2}}\] \[-0.1\,x\,dx\,=K{{E}_{F}}-500\] \[-0.1\int_{20}^{30}{x\,}dx=K{{E}_{F}}-500\] \[-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{20}^{30}=K{{E}_{F}}-500\] \[-\frac{0.1}{2}[{{(30)}^{2}}-{{(20)}^{2}}]=K{{E}_{F}}-500\] \[-\frac{0.1}{2}[900-400]=K{{E}_{F}}-500\] \[=25=K{{E}_{F}}-500\] Final kinetic energy, \[K{{E}_{F}}=475J\]You need to login to perform this action.
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