Chapter 6, Problem 75GP

To determine

a) The speed of the mass when it is at the normal length

b) The speed of the mass at half its original extension

Answer to Problem 75GP

Solution:

a) The speed is $v=\sqrt{\frac{Fx}{m}}$ .

b) The speed is $v=\sqrt{\frac{3Fx}{4m}}$ .

Explanation of Solution

Given

The mass m is attached to a spring and held stretched by a force F .

The stretch by the force F is x .

Therefore, the spring constant $k=\frac{F}{x}$

Formula used

Energy conservation conditions must be used here.

For part a, the kinetic energy of the mass is equal to the spring potential energy, $\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$ .

For part b, the spring potential energy, $\frac{1}{2}k{x}^2$ is equal to the kinetic energy, $\frac{1}{2}m{v}^2$ and spring potential energy $\frac{1}{2}k{\left(\frac{x}{2}\right)}^2$ , since $x/2$ is the half of original extension.

Calculations

Part a

  $\begin{array}{l}\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2\\ v^2=\frac{k{x}^2}{m}\\ \because k=\frac{F}{x}\\ v^2=\frac{Fx}{m}\\ v=\sqrt{\frac{Fx}{m}}\end{array}$

Therefore, the speed for return to its normal length is $v=\sqrt{\frac{Fx}{m}}$ .

Part b

For return to half the original extension,

  $\begin{array}{l}\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{\left(\frac{x}{2}\right)}^2\\ \frac{1}{2}k{x}^2\left(1-\frac{1}{4}\right)=\frac{1}{2}m{v}^2\\ k{x}^2\times \frac{3}{4}=m{v}^2\\ \frac{F}{x}{x}^2\times \frac{3}{4}=m{v}^2\\ m{v}^2=\frac{3Fx}{4}\\ v^2=\frac{3Fx}{4m}\\ v=\sqrt{\frac{3Fx}{4m}}\end{array}$

Conclusion: Therefore, the speed for return to its half of original extension is $v=\sqrt{\frac{3Fx}{4m}}$ .