Question

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the

spring is 250 N/m. What is the potential energy of the spring?

Answer 1

31.25 hope this helps:)

Answer 2

31.25

Step-by-step explanation:

The formula for the potential energy of a spring is
`(1)/(2)kx^2`, where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get
`(1)/(2)\cdot 250\cdot 0.25=31.25`. Hope this helps!