Question

no links or i will report. What is the potential energy of stretched spring, if the spring constant is 40 N/m and the elongation is 5 cm?

Answer 1

Answer: 0.05 J

Explanation:

List all variables before solving:

K = 40 N/m

x = 5 cm (convert to 0.05 m)

Use the equation for potential energy (PE):

PE =
`(1)/(2)` k
`x^(2)`

Plug in the given variables and solve:

PE = (
`(1)/(2)` ) (40 N/m) (0.05 m)^2

The answer is 0.05 J