Question

If a spring is stretched to twice the lengtj of its equilibrium position, by what factor does the energy stored in the spring change?

Answer 1

Answer:

the energy increases 4 times

Explanation:

A spring has an elastic potential energy that is given by the expression

          K_e = ½ K (x-x₀)²

where x is the distance from the equilibrium point and k is the return constant

if the spring is stretched at  x-x₀ = 2x₀, the energy value

        K_e = ½ k (2x₀)²

        K_e = ½ k 4 x₀²

        K_e = 4 (½ k x₀²)

        $\frac{K_e}{ \frac{1}{2} k x_o^2}$ = 4

therefore the energy increases 4 times