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

Step-by-step explanation:

PE =
`(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=(1)/(2)(2.5)(.2^2)` which gives us

PE = .05 J