Chapter 6.6, Problem 7E

(a).

To determine

To Calculate: The work required to stretch a spring.

Answer to Problem 7E

The work required to stretch the spring from to is approximately $1.04J$

Explanation of Solution

Given Information: The natural length of the spring is and the work done in stretching it to is .

Method 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.

Therefore, the work done in stretching a spring of spring constant $k$ from elongation length $x_1$ to $x_2$ is

  $W={{\displaystyle \int }}_{x_1}^{x_2}kx\cdot dx={\frac{{\text{kx}}^2}{2}]}_{{\text{x}}_1}^{{\text{x}}_2}$

  $\Rightarrow W=\frac{1}{2}\text{k}\left({\text{x}}_2{}^2-{\text{x}}_1{}^2\right)$

Calculation: Put the given values $W=2, x_1=30-30=0$ and in the work equation for a spring to write

  $2=\frac{1}{2}k\left({0.12}^2-0^2\right)$

  $\Rightarrow 2=\frac{1}{2}k\left(0.0144\right)$

  $\Rightarrow \underset{\_}{k \approx 277.8}$

Now, to find the work required to stretch the spring from to , put the above calculated value of $k$ , the values andin the work equation for a spring to write

  $W\approx \frac{1}{2}\times 277.8\times \left({0.1}^2-{0.05}^2\right)$

  $\approx \frac{1}{2}\times 277.8\times \left(0.0075\right)$

(b).

To determine

To Calculate: The elongation in a spring given its spring constant and the force on it.

Answer to Problem 7E

Explanation of Solution

Given Information: The force is $30N$ and the spring constant from the previous sub-part is approximately

Method Used: Hooke’s Law states that for a spring, $f\left(x\right)=kx$ , 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.

Calculation: Put the values of $f=30 N$ and $k\approx 277.8$ in equation of Hooke’s Law to write

  $30\approx 277.8 \times x$

  

Therefore, the elongation for a force of $30N$ is approximately 10.8cm