Question

A spring has a spring constant of 120 N/m. How much potential energy is stored in the spring as it is stretched 0.2 m?

1) 2.4 J
2) 12 J
3) 4.8 J
4) 24 J

Answer 1

The potential energy stored in a spring with a constant of 120 N/m stretched 0.2 m is 2.4 J, calculated using the formula PE = (1/2)kx^2.

Step-by-step explanation:

The question involves calculating the potential energy stored in a spring with a known spring constant as it is stretched. According to Hooke's law, the potential energy (PE) stored in a spring stretched or compressed by a distance x from its equilibrium position is given by the equation PE = (1/2)kx2, where k is the spring constant and x is the displacement.

In this case, the spring constant k is given as 120 N/m, and the displacement x is 0.2 m. Plugging these values into the formula:

PE = (1/2)(120 N/m)(0.2 m)2 = (1/2)(120)(0.04) = 2.4 J.

Therefore, the amount of potential energy stored in the spring is 2.4 J when it is stretched 0.2 m.