Question

A mass of 0.368 kilograms is suspended from a spring, and the spring stretches 12.3 centimeters. What is the spring constant of the spring? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

Given data:

* The mass of the given body is 0.368 kg.

* The spring is stretched to the length 12.3 cm.

Solution:

The diagrammatic representation of the given case is,

The restoring force acting on the spring is same as the weight of the body acting vertically downward.

`F=w`

The weight of the body is,

`w=mg`

where m is the mass of the body, g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} w=0.368*9.8 \\ w=3.61\text{ N} \end{gathered}`

The restoring force acting on the spring is,

`F=kx`

where k is the spring constant and x is the displacement of the spring from an equilibrium,

Substituting the known values,

`\begin{gathered} 3.61=k*12.3*10^(-2) \\ k=(3.61)/(12.3*10^(-2)) \\ k=(3.61)/(0.123) \\ k=29.3\text{ N/m} \end{gathered}`

Thus, the spring constant of the spring is 29.3 N/m.