Chapter 8.1, Problem 29E

a. $$

To determine

Tofindthe force which will be required to stretch the spring $9cm$ beyond its natural length.

Answer to Problem 29E

  $18\text{Newton}$

Explanation of Solution

Given information:

A certain spring requires a force of $6Newton$ to stretch it $3cm$ beyond its natural length.

Formula:

Hooke’s Law for springs states that the force it takes to stretch or compress a string $x$ units from its natural length is given by

  $F=kx$ where $k$ -force constant of a spring

Hooke’s Law for springs states that the force it takes to stretch or compress a string $x$ units from its natural length is given by

  $F=kx$ where $k$ -force constant of a spring

Given that the spring takes $6Newton$ to stretch the spring $3cm$ beyond its natural length, then the force constant is given by

  $k=\frac{6N}{3cm}=2N/cm$

Stretching the spring $9cm$ beyond its natural length requires a force of,

  $\begin{array}{l}F=kx\\ F=2\times 9\\ F=18N\end{array}$

Therefore, 18 Newton of force is required to stretch the spring.

b. $$

To determine

To find thework done in stretching the spring $9cm$ beyond its natural length.

Answer to Problem 29E

The work done in stretching the spring is $81\text{N}\cdot \text{cm}$ .

Explanation of Solution

Given information:

A certain spring requires a force of $6\text{ Newton}$ to stretch it $3cm$ beyond its natural length.

The work done in stretching the spring $9cm$ beyond its natural length can be found by solving the integral:

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{9}{\int }}2xdx}={\left[x^2\right]}_0^9\\ =81\text{N}\cdot \text{cm}\end{array}$

Therefore, thework done in stretching the spring is $81\text{N}\cdot \text{cm}$ .