The steel spring is stretched by a force of 50 N. When the spring is stretched with an additional force of 80 N

The steel spring is stretched by a force of 50 N. When the spring is stretched with an additional force of 80 N, it is extended by another 20 cm. What work is done with this extension?

Let’s translate the amount of spring elongation from centimeters to meters:

20 cm = 0.2 m.

First, we find the stiffness of the steel spring, deriving it from the formula for the elastic force:
Fcont = kx ->

k = Fcont / x = 80 / 0.2 = 400 n / m.

Knowing the stiffness of the spring, we find its elongation in the first case, also deriving it from the formula for the elasticity of the body:

Fcont = kx ->

x1 = Fcont / k = 50/400 = 0.125 m.

The work of the spring is equal to the difference in its kinetic energies:

A = Ek2 – Ek1 = (kx2 ^ 2 – kx1 ^ 2) / 2 = k (x2 ^ 2 – x1 ^ 2) / 2 = 400 (0.2 ^ 2 – 0.125 ^ 2) / 2 = 200 (0, 04 – 0.015625) = 200 * 0.024375 = 4.875 J.

Answer: 4.875 J.