Question

A spring with a spring constant of 125 n/m is stretched 2. 0 cm from its equilibrium position. What is the change in the spring potential energy?.

Answer 1

Answer: 0.025 J or 25 mJ

Step-by-step explanation:

Elastic potential energy = 1/2 * k * x^2

Where k is the spring constant (N/m) and x is the springs stretch from the point of equilibrium.

=1/2*125N/m * (0.02m)^2 = 0.025 J or 25mJ