Question

The spring constant of a spring is 25 N/m. The spring is extended by 50 cm. How much elastic potential energy is stored? a) 0.0625 J b) 0.125 J c) 0.25 J d) 0.5 J

Answer 1

To calculate the elastic potential energy stored in a spring extended by 50 cm with a spring constant of 25 N/m, use the formula U = 1/2 k x^2. By plugging in the values, we find that the energy stored is 0.25 J.

Step-by-step explanation:

The question involves calculating the amount of elastic potential energy stored in a spring when it is extended by a certain distance. The formula to calculate the elastic potential energy (U) stored in a spring is U = 1/2 k x^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position, in meters. Given the spring constant of 25 N/m and an extension of 50 cm (0.50 m), the elastic potential energy can be calculated as follows:

U = 1/2 (25 N/m) (0.50 m)^2

U = 1/2 (25) (0.25)

U = 1/2 (6.25)

U = 3.125 J

Therefore, the correct answer is c) 0.25 J.