Question

In her garage, a bicyclist hangs a box of bike parts from a long spring attached to the ceiling. The spring stretches 0.80 m with the box attached and has a spring constant of 150 N/m. What is the elastic potential energy stored in the spring? A. –60 J B. –48 J C. 48 J D. 60 J

Answer 1

Using the formula for elastic potential energy, the spring with a constant of 150 N/m stretched by 0.80 m stores 48 J of energy, corresponding to option C.

Step-by-step explanation:

The elastic potential energy stored in a spring can be calculated using the formula U = (1/2)kx², where U is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. In this case, the spring constant k is 150 N/m and the displacement x is 0.80 m. Plugging these values into the formula gives us:

U = (1/2)(150 N/m)(0.80 m)²

U = (1/2)(150 N/m)(0.64 m²)

U = (1/2)(96 J)

U = 48 J

Therefore, the elastic potential energy stored in the spring is 48 J, which corresponds to option C.