What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 6.0 cm from its relaxed length?

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The equation for Hooke's Law is:

F = -kx

Where:
F is the force applied to the spring
k is the spring constant
x is the displacement from the equilibrium position

We're given that the spring stores 25 J of elastic potential energy, which is the work done to compress the spring. The potential energy stored in a spring is given by:

PE = (1/2)kx^2

Let's rearrange the equation to solve for the spring constant:

k = 2PE / x^2

Substituting the given values:
PE = 25 J
x = 6.0 cm = 0.06 m

k = 2(25 J) / (0.06 m)^2

Now, we can calculate the spring constant.