Problem: Hooke's law states that the force on a spring varies directly with the distance that it is stretched. If a spring has a k value of 100 newtons per meter and it is stretched 0.50 meters, what is the restoring force of the spring?

Respuesta :

Answer:

Restoring force of the spring is 50 N.

Explanation:

Given that,

Spring constant of the spring, k = 100 N/m

Stretching in the spring, x = 0.5 m

We need to find the restoring force of the spring. It can be calculated using Hooke's law as "the force on a spring varies directly with the distance that it is stretched".

[tex]F=kx[/tex]

[tex]F=100\ N/m\times 0.5\ m[/tex]

F = 50 N

So, the restoring force of the spring is 50 N. Hence, this is the required solution.