Question

A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the potential energy stored in the spring?
____ J

Answer 1

Answer:

113.4 J

Explanation:

Elastic Potential Energy

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is

Δx = 2.5 m - 0.7 m = 1.8 m

The energy stored in the spring is:

$\displaystyle PE = \frac{1}{2}70(1.8)^2$

PE = 113.4 J