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

Answer 2

The energy stored in the spring changes by a factor of 4

The formula for calculating the energy stored in the spring is expressed as:

$E=\frac{1}{2}ke^2$

If the spring is stretched to twice the length of its equilibrium position

$E_2=\frac{1}{2}(2k)^2\\E_2=\frac{1}{2}4k^2$

Take the ratio of the energy stored in the spring

$\frac{E_2}{E} =\frac{1/2(4kx^2)}{1/2kx^2}\\ \frac{E_2}{E} =\frac{4}{1}\\E_2 = 4E$

This shows that the energy stored in the spring changes by a factor of 4

Learn more here: https://brainly.com/question/24555300