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?​

Respuesta :

22.5 J

Explanation:

Given:

x = 3 m

[tex]k = 5\:\text{N/m}[/tex]

The spring potential energy [tex]PE_s[/tex] is

[tex]PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2[/tex]

[tex]\:\:\:\:\:\:\:=22.5\:\text{J}[/tex]