Physics
JAMB 1991
A spring of force constant 1500Nm
-1 is acted upon by a constant force of 75N. Calculate the potential energy stored in the string
-
A.
1.9J
-
B.
2.8J
-
C.
3.45J
-
D.
0.43J
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
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