Chapter 6.6, Problem 6E

To determine

To Calculate: The work required to stretch a spring.

Answer to Problem 6E

The work required to stretch the spring to $25cm$ is $0.3215N$

Explanation of Solution

Concept Used:

• The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

• Hooke’s Law: $f\left(x\right)=kx$ , where $x$ is the amount by which the spring is elongated, $k$ is the spring constant and $f\left(x\right)$ is the force required to maintain the spring at that elongation length.

Given Information: The natural length of the spring is $20cm$ . If the stretched length is $30 cm$ , then the force required is .

Calculation:The elongation of the spring is

  $x=\text{stretched length}-\text{natural length}=30-20=10cm=0.1m$

Put the given value of force and the above calculated elongation in Hooke’s Law to write

  $f\left(0.1\right)=k\times 0.1$

  $\Rightarrow 25= k\times 0.1$

  $\Rightarrow k = 250$

So, the force is given by $f\left(x\right)=250x$ .

Now for a stretched length of $25 cm$ , elongation is $25-20=5cm=0.05m$ .

So, put the above expression $f\left(x\right)=250x$ and $x_1=0$ (at natural length) and $x_2=0.05$ in the equation for work to write

  $W={{\displaystyle \int }}_0^{0.05}250xdx$

  $={\frac{250x^2}{2}]}_0^{0.05}$

  $=\left(\frac{250\times 0.0025}{2}\right)-0$