Question

A ball of mass 1.95 kg is hung from a spring which stretches a distance of 0.4300 m. The spring constant of the spring (a scalar quantity) is ?? N/m.

Answer 1

The spring constant of the spring is 44.44 N/m.

Step-by-step explanation:

Hooke's Law states that the force F needed to extend or compress a spring by some distance X is proportional to that distance.

The equation for this is F = kX, where k is the spring constant and X is the extension (or compression) of the spring.

In this scenario, a mass of 1.95 kg causes the spring to stretch 0.4300 m.

The force applied to the spring is due to the weight of the mass and can be found by multiplying the mass by the acceleration due to gravity (9.8 m/s2).

Therefore, F = 1.95 kg * 9.8 m/s2

= 19.11 N.

Now, we can rearrange Hooke's Law to solve for k: k = F / X.

Substituting the given values in, k = 19.11 N / 0.4300 m

k= 44.44 N/m.