JAMB - Physics (1991 - No. 6)
A spring of force constant 1500Nm-1 is acted upon by a constant force of 75N. Calculate the potential energy stored in the string
1.9J
2.8J
3.45J
0.43J
Paaiškinimas
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