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=(mg)/(x)`

On substituting the known values,

`\begin{gathered} k=(35*9.8)/(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