Question

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3 meters.

What is the SPE?​

Answer 1

22.5 J

Step-by-step explanation:

Given:

x = 3 m

`k = 5\:\text{N/m}`

The spring potential energy
`PE_s` is

`PE_s = (1)/(2)kx^2 = (1)/(2)(5\:\text{N/m})(3\:\text{m})^2`

`\:\:\:\:\:\:\:=22.5\:\text{J}`