Question

A spring with a spring constant of 2.5 N/m is stretched by 0.2 m. What is the potential energy in the spring

Answer 1

Answer:

Explanation:

PE = $\frac{1}{2}k$(Δ$x^2$)

where k is the spring constant and Δx is the displacement of the spring when an object is hung from it. Plugging in:

$PE=\frac{1}{2}(2.5)(.2^2)$ which gives us

PE = .05 J