Question

Listena spring of negligible mass with a spring constant of 200 newtons per meter is stretched 0.2 meter. how much potential energy is stored in the spring?

O 40 J
O 20 J
O 8 J
O 4 J

Answer 1

The potential energy stored in a spring with a spring constant of 200 newtons per meter and stretched 0.2 meter is calculated using the formula PE = (1/2)kx², resulting in 4 joules.

Step-by-step explanation:

The question is about calculating the potential energy stored in a spring. The formula for potential energy (PE) in a spring is given by PE = (1/2)kx², where k is the spring constant, and x is the displacement from the spring's equilibrium position. Given the spring constant (k) of 200 newtons per meter and the displacement (x) of 0.2 meter, we can substitute these values into the formula to get the potential energy stored.

PE = (1/2)(200 N/m)(0.2 m)² = (1/2)(200)(0.04) J

PE = (1/2)(8) J = 4 J

So, the potential energy stored in the spring is 4 joules (4 J).