Question

A spring has a force constant 18 N/m. If it is compressed 50 cm from its equilibrium length, how much mass will it have gained?

Short answer

The gain in mass of spring is$2.5\times {10}^{-17}\text{kg}$

Step-by-step solution

Identification of given data

The force constant of spring is $k=18\text{N}/\text{m}$

The compressed length of spring is $x=50\text{cm}$

Elastic Potential Energy

The energy stored in an elastic material due to variation in the equilibrium position of atoms is called the elastic potential energy. This energy of spring in the problem is equated to rest energy of spring to find the gain in mass.

Determination of gain in mass of spring

The gain in mass of spring is given as:

$\Delta m=\frac{k{x}^2}{2c^2}$

Here, c is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$ ,

Substitute all the values in the equation.

$$\begin{gathered} \Delta m=\frac{\left(18\text{N}/\text{m}\right){\left[\left(50\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right]}^2}{2{\left(3\times {10}^8\text{m}/\text{s}\right)}^2} \\ \Delta m=0.25\times {10}^{-16}\text{kg} \\ \Delta m=2.5\times {10}^{-17}\text{kg} \end{gathered}$$

Therefore, the gain in mass of spring is $2.5\times {10}^{-17}\text{kg}$.