Question

Hector stretches a spring with a spring constant of 3 N/m until it is extended by 50 cm. What is the elastic potential energy stored by the spring?

Answer 1

The elastic potential energy stored in the spring is 0.375 J.

The formula for elastic potential energy is:

E = 1/2 * k * x^2

where:

* E is the elastic potential energy in Joules

* k is the spring constant in N/m

* x is the distance the spring is stretched or compressed from its equilibrium position in meters

In this problem, we have:

* k = 3 N/m

* x = 0.5 m (50 cm)

Substituting these values into the formula, we get:

E = 1/2 * 3 * 0.5^2 = 0.375 J

Therefore, the elastic potential energy stored in the spring is 0.375 J.