A) 1.5 Joule
B) 2.0 Joule
C) 2.5 Joule
D) 3.0 Joule
Correct Answer: C
Solution :
Force constant of a spring \[k=\frac{F}{x}=\frac{mg}{x}=\frac{1\times 10}{2\times {{10}^{-2}}}\,\,\,\Rightarrow \,\,k=500\,N/m\] Increment in the length \[= 60 - 50 = 10 cm\] \[\operatorname{U}=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}\,\,500{{(10\times 1{{0}^{-2}})}^{2}}=2.5\,J\]You need to login to perform this action.
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