Question

A spring stores potential energy U0 when it is compressed a distance x0 from its uncompressed length.

Answer 1

Answer:

Explanation:

Using Hooke's law

Elastic potential energy = 1/2 K x²

K is elastic constant of the spring

x is the extension of the spring

a) The elastic potential energy when the spring is compressed twice as much  Uel = 1/2 k (2x₀) ² =  4 (1/2 kx₀²)= 4 U₀

b) when is compressed half as much Uel = 1/2 k $\frac{x0}{4} ^{2}$ = $\frac{1}{4}$ ( U₀)

c) make x₀ subject of the formula in the equation for elastic potential

      x₀ =$\sqrt{\frac{2U0}{K} }$

x, the amount it will compressed to tore twice as much energy = $\sqrt{\frac{2 (2U0)}{K} }$

x = √2 x₀

d) x₁, the new length it must be compressed to store half as much energy = $\sqrt{\frac{2 (\frac{1}{2})U0 }{K} }$

x₁ = $\sqrt{\frac{1}{2} }$ x₀