A) 4.900 joule
B) 2.450 joule
C) 0.495 joule
D) 0.245 joule
Correct Answer: B
Solution :
\[K=\frac{F}{x}\]\[=\frac{40}{2\times {{10}^{-2}}}=0.2\ N/m\] Work done\[=\frac{1}{2}K{{x}^{2}}=\frac{1}{2}\times (0.2)\times {{(0.05)}^{2}}=2.5\ J\]You need to login to perform this action.
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