Question

A spring has a spring constant of 120 newtons per meter. Calculate the elastic potential energy stored in the spring when it is stretched 2.0 centimeters.

Answer 1

The elastic potential energy stored in a spring with a spring constant of 120 N/m when stretched by 2.0 centimeters is 0.024 joules.

Step-by-step explanation:

To calculate the elastic potential energy stored in a spring when stretched, we use the formula:

PE = 1/2 kx²

Where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position in meters. In this case, the given spring constant is 120 newtons per meter (N/m), and the displacement is 2.0 centimeters, which needs to be converted to meters to use in the formula:

x = 2.0 cm = 0.02 m

Applying the given values to the formula:

PE = 1/2 (120 N/m)(0.02 m)²

This results in:

PE = 0.5 × 120 N/m × (0.02 m)²PE = 0.5 × 120 × 0.0004PE = 0.024 J

Therefore, the elastic potential energy stored in the spring when it is stretched 2.0 centimeters is 0.024 joules (J).

Answer 2

The elastic potential of a spring can be calculated using the formula
E = (1/2) k x²
where are given that
k = 120 N/m
the stretched length is
x = 0.02 m

The elastic potential is
E = (1/2) (120 N/m) (0.02 m)²
E = 0.024 N-m