Question on: JAMB Physics - 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
Ask EduPadi AI for a detailed answer
Correct Option: A
P.E = \(\frac{1}{2}\)Fe
= \(\frac{1}{2}\)\(\frac{F}{K}\)
= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)
= 1.9J
= \(\frac{1}{2}\)\(\frac{F}{K}\)
= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)
= 1.9J
Add your answer
Please share this, thanks!
No responses