Q23 DQ - University Physics with Modern Physics

Question

When a certain force is applied to an ideal spring, the spring stretches a distance x from its unstretched length and does work W. If instead twice the work is applied, what distance (in terms of x) does the spring stretch from its unstretched length, and how much work (in terms of W) is required to stretch it this distance?

Step-by-Step Solution

Answer

$\mathrm{x}=2\sqrt{\frac{\mathrm{F}}{\mathrm{k}}}\mathrm{and} \mathrm{W} =\sqrt{\frac{\mathrm{F}}{\mathrm{k}}}$

Step 1: About the spring force

The spring force is given by: $\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$

Here, k is the spring constant and x is the distance.

Step 2: Determination of the distance and work required to stretch the distance

It is given that twice the work is applied.

$\mathrm{F}=\frac{1}{2}{\mathrm{kx}}^2$ 

This can be rearranged to,

 $\mathrm{x}=\sqrt{\frac{2\mathrm{F}}{\mathrm{k}}}$ 

Now, when the force is twice 

$$\begin{gathered} 2\mathrm{F}=\frac{1}{2}{\mathrm{kx}}^2 \\ \frac{4\mathrm{F}}{\mathrm{k}}={\mathrm{x}}^2 \end{gathered}$$ 

  Therefore, x = 2$\sqrt{\frac{\mathrm{F}}{\mathrm{k}}}$

Now, the work required to stretch the distance is given by 

Work = Force x Distance

 

$$\begin{gathered} \mathrm{W}=\frac{1}{2}{\mathrm{kx}}^2\times 2\sqrt{\frac{\mathrm{F}}{\mathrm{k}}} \\ \mathrm{W}={\mathrm{kx}}^2\sqrt{\frac{\mathrm{F}}{\mathrm{k}}} \end{gathered}$$

Thus, the required distance is $\mathrm{x}=2\sqrt{\frac{\mathrm{F}}{\mathrm{k}}}$ and required work done is $\mathrm{W}={\mathrm{kx}}^2\sqrt{\frac{\mathrm{F}}{\mathrm{k}}}$