Work Energy Power

NEET level, single correct, medium

2 of 183

Question 2 of 183

A spring of force constant k=400 N/mk = 400\,\text{N/m} is compressed by 0.1 m0.1\,\text{m}. The elastic potential energy stored is:

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Answer: (D) 2 J2\,\text{J}

U=12kx2=0.5×400×(0.1)2=2 JU = \frac{1}{2} k x^2 = 0.5 \times 400 \times (0.1)^2 = 2\,\text{J}.

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