Question

A block with a mass of 0.6 kg is attached to a spring and it descends with an elongation of 0.15m. What will be the constant of the spring?

Answer 1

The mass of the block. m=0.6 kg

The displacement of the spring, x=0.15 m

Total force applied on the spring from the mass is given by,

`F=mg\text{ }\rightarrow\text{ (i)}`

Where g is the acceleration due to gravity.

The restoring force of the spring is given by,

`F=kx\text{ }\rightarrow\text{ (ii)}`

Where k is the spring constant

Thus, from equation (i) and equation (ii),

`\begin{gathered} mg=kx \\ \Rightarrow k=(mg)/(x) \end{gathered}`

On substituting the known values,

`\begin{gathered} k=(0.6*9.8)/(0.15) \\ =39.8\text{ N/m} \end{gathered}`

Thus the force constant of the spring is 39.8 N/m