Question

A 35 kg weight displaces a spring from equilibrium by 65 cm. What is the spring constant in N/m?

Answer 1

Given,

The mass, m=35 kg

The displacement of the spring from its equilibrium position, x=65 cm=0.65 m

The magnitude of the restoring force of the spring is given by,

$F=kx$

And the restoring force will be equal to the weight due to the mass attached to the spring.

Therefore,

$F=mg=kx$

Where g is the acceleration due to gravity.

On rearranging the above equation,

$k=\frac{mg}{x}$

On substituting the known values,

$\begin{gathered} k=\frac{35\times9.8}{0.65} \\ =527.7\text{ N/m} \end{gathered}$

Thus the value of the spring constant of the given spring is 527.7 N/m