Question

How many joules of energy can you store in a spring whose stiffness is, 0.6 N/m by starting from a relaxed spring and stretching it 20cm?

Short answer

A spring having0.6 N/m stiffness and starting to stretch it from a relaxed position upto 20 cm can store0.012 J of energy.

Step-by-step solution

Given Information

Stiffness of spring K=0.6 N/m,length of spring stretched $=20\times {10}^{-2}\mathrm{m}$.

Definition of Potential Energy

The energy owned by a body as a result of its position, shape, or change of configuration is called as potential energy.

The potential energy is given by,

$U_s=\frac{1}{2}K{s}^2$

Here k is spring constant and sis distance from equilibrium

Energy calculation for the spring

In this case, all that’s need to be done is apply the concept of elastic potential energy to figure out how much energy would be stored in a spring with the specified properties.

Substitute $k=0.6\mathrm{N}/\mathrm{m}\mathrm{and}\mathrm{s}=20\times {10}^{-2}\mathrm{m}$into the formula of potential energy.

$$\begin{gathered} U_{s_{}}=\frac{1}{2}\left(0.6\mathrm{N}/\mathrm{m}\right){\left(20\times {10}^{-2}\mathrm{m}\right)}^2\left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =0.012\mathrm{J} \end{gathered}$$

Therefore,$0.012\mathrm{J}$ of energy can be stored in a spring having stiffness of 0.6 N/m.