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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