A spring of force constant 1500Nm-1 is acted upon by a constant force of 75N. Calculate the potential energy stored in the string
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Correct Answer: Option A
Explanation:
P.E = \(\frac{1}{2}\)Fe
= \(\frac{1}{2}\)\(\frac{F}{K}\)
= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)
= 1.9J
P.E = \(\frac{1}{2}\)Fe
= \(\frac{1}{2}\)\(\frac{F}{K}\)
= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)
= 1.9J