To compress the spring by 5 cm, a force of 15 N is applied. what is the energy of the deformed spring

Given:

F = 15 Newton – force applied to deform the spring;

dx = 5 centimeters = 0.05 meters – the amount of spring deformation.

It is required to determine E (Joule) – the energy of the deformed spring.

According to Hooke’s law, we find the coefficient of spring stiffness:

k = F / dx = 15 / 0.05 = 300 N / m.

Then the energy of the deformed spring will be equal to:

E = k * dx ^ 2/2 = 300 * 0.05 ^ 2/2 = 150 * 0.05 ^ 2 = 0.0025 * 150 = 0.375 Joules.

Answer: The energy of the deformed spring is 0.375 Joules.