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 =
( U₀)
c) make x₀ subject of the formula in the equation for elastic potential
x₀ =
x, the amount it will compressed to tore twice as much energy =
x = √2 x₀
d) x₁, the new length it must be compressed to store half as much energy =
x₁ = x₀