Question

A spring of force constant 1200Nm is acted on by a constant force of 60N. Calculate the potential energy stored in the spring?

Answer 1

Let's first find the elongation of the spring from the Hooke's Law:

$$F=kx,$$ $$x=\dfrac{F}{k}=\dfrac{60\ N}{1200\ \dfrac{N}{m}}=0.05\ m.$$

Finally, we can calculate the potential energy stored in the spring:

$$PE=\dfrac{1}{2}kx^2=\dfrac{1}{2}\cdot 1200\ \dfrac{N}{m}\cdot (0.05\ m)^2=1.5\ J.$$

Answer:

$PE=1.5\ J.$