Question

How far must a spring with a constant of 3.0 N/m be pulled back to apply 2.0 N of force on the 0.50 kg mass that is attached to it?

Answer 1

0.67 meters (or 67 centimeters)

Step-by-step explanation:

F = -kx

Where:

F = force exerted by the spring (in newtons, N)

k = spring constant (in newtons per meter, N/m)

x = displacement of the spring from its equilibrium position (in meters, m)

Given:

k = 3.0 N/m (spring constant)

F = 2.0 N (force exerted on the mass)

m = 0.50 kg (mass)

We can rearrange Hooke's Law to solve for the displacement x:

x = -F / k

Plugging in the given values:

x = -2.0 N / 3.0 N/m

x = -0.67 m

So, the spring must be pulled back by a distance of 0.67 meters (or 67 centimeters) in order to apply a force of 2.0 N on the 0.50 kg mass attached to it. Note that the negative sign indicates that the spring is being stretched or pulled in the opposite direction of its equilibrium position.