Chapter 6.6, Problem 5E

To determine

To Calculate: The work required to stretch a spring.

Answer to Problem 5E

The work done to stretch the spring by $6in$ is $45 ft\cdot lb$

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 elongation of the spring is $4 in$ and the force required is $10 lb$ .

Calculation: Put the given values of force and elongation in Hooke’s Law to write

  $f\left(4\right)=k\cdot 4$

  $\Rightarrow 10 = k\cdot 4$

  $\Rightarrow k = 2.5$

So, the force is given by $f\left(x\right)=2.5x$

Put the above expression $f\left(x\right)=2.5x$ , $x_1=0$ (at natural length) and $x_2=6$ in the equation for work to write

  $W={{\displaystyle \int }}_0^{6}2.5xdx$

  $={\frac{2.5x^2}{2}]}_0^6$

  $=\left(\frac{2.5\times 36}{2}\right)-0$

  $=45 ft\cdot lb$