Question

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?

Short answer

Answer

To increase its length from 5 cm to 9 cm,- 0.04 J amount of work must be done.

Step-by-step solution

Definition of Work

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.

Given Information 

The relaxed strength of the spring is [ = 6 cm

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

Finding Change in length when length is compressed 5 cm

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

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

Convert the unit from cm to m.

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

Finding Change in length when length is extended to 9 cm 

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.

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

 Step 5: Finding Potential energy due to compression

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.

Finding Potential energy due to expansion

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

Finding the Work done

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.