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) 4.8 J
2) 2.4 J
3) 12 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 calculated using the formula PE = 1/2 k x², resulting in 2.4 J.

Step-by-step explanation:

To determine the potential energy stored in a spring, we need to use the formula for elastic potential energy, which is given as PE = 1/2 k x², where k is the spring constant and x is the displacement of the spring from its rest position.

In this problem, the spring has a spring constant of 120 N/m and it has been stretched by 0.2 m. Using the formula, we calculate the potential energy as follows:

The potential energy stored in a spring can be calculated using the equation ΚPE = 1/2kx², where k is the spring constant and x is the displacement of the spring. In this case, the spring constant is 120 N/m and the displacement is 0.2 m. Plugging in these values, we get:

PE = 1/2(120 N/m)(0.2 m)² = 1/2(120 N/m)(0.04 m²) = 2.4 J.

Therefore, the correct answer is 2.4 J.