Question

Simple H...

If a mass attached to a vertical spring stretches the spring 2.0 m with a spring
constant of 5 N/m from its original equilibrium position, what is the force acting
on the string?

Answer 1

10N

Step-by-step explanation:

The force acting on the spring can be found using Hooke's Law:

F = kx

where F is the force, k is the spring constant, and x is the displacement from equilibrium.

Plugging in the given values:

F = (5 N/m)(2.0 m)

F = 10 N

Therefore, the force acting on the spring is 10 N.