Given that,
The force constant of a spring, k = 570.8 N/m
It is stretched 4.12 cm from the equilibrium.
To find,
The potential energy stored in the spring.
Solution,
The potential energy stored in the spring is given by the formula as follows :
[tex]E=\dfrac{1}{2}kx^2[/tex]
Put all the given values,
[tex]E=\dfrac{1}{2}\times 570.8 \times (4.12\times 10^{-2})^2\\\\E=0.484 J[/tex]
So, the required potential energy is 0.484 J.
