Question: A spring has a relaxed length of 6 cm and a stiffness of How much work must you do to change its length from 5cm to 9cm?

When an object is pushed from its original location by an external force, it is said to be doing work.

The work done as a result of the force is given by W = F . S or W=$F_{XCOS\theta }$. Where is the work done, is the external force, and is the displacement.

The relaxed strength of the spring is [ = 6 cm

The stiffness of the spring is K_S= 100 N/m

Subtract 5 cm compressed length to find for the change in length.

$$\begin{gathered} \Delta l=l_1-l_2 \\ =6\text{cm}-5\text{cm} \\ =1cm \end{gathered}$$

$$\begin{gathered} ∆I=1cm\left(\frac{1cm}{100cm}\right) \\ ={10}^{-2}m \end{gathered}$$

The spring's original length is and the spring's given extended length is .

Subtract from to find the change in length.

$$\begin{gathered} \Delta l=l_1-l_2 \\ =9\text{cm}-6\text{cm} \\ =3cm \end{gathered}$$

Convert the unit from cm to m.

$$\begin{gathered} ∆I=3cm\left(\frac{1cm}{100cm}\right) \\ =3\times {10}^{-2}m \end{gathered}$$

The spring's potential energy due to compression can be stated as$U_1=\frac{1}{2}{k}_s{\left(\Delta l\right)}^2$.

Substitute and into the formula of potential energy to get the potential energy when the length is compressed to 5 cm.

Substitute and into the formula of potential energy to get the potential energy when the length is extended to 9 cm.

The work done by the spring is simply equal to the change in potential energy. It can be stated .

Substitute the values into the formula to find the work done.