A) 475J
B) 450J
C) 275J
D) 250J
Correct Answer: A
Solution :
Force =-0.1 x J/m Let dx be the small displacement \[\operatorname{dW} = F.dx\] Integrating both sides. \[\int\limits_{20}^{30}{\operatorname{dw}}=\int\limits_{20}^{30}{\operatorname{F}.dx}\] \[\operatorname{W}=\int\limits_{20}^{30}{-0.1x.\,dx}\] \[{{\operatorname{K}}_{f}}-{{K}_{i}}=-0.1{{\left[ \frac{{{x}^{2}}}{2} \right]}^{20}}_{30}\] \[{{\operatorname{K}}_{f}}-\frac{1}{2}\times 10\times {{\left( 10 \right)}^{2}}=-0.1{{\left[ {{x}^{2}} \right]}^{30}}_{20}\] \[{{\operatorname{K}}_{f}}-500=-\frac{1}{2}\times \left[ 900-400 \right]\] \[{{\operatorname{K}}_{f}}-500=-25\] \[{{\operatorname{K}}_{f}}=475\operatorname{J}\]
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