Chapter 7.1, Problem 35E

(a)

To determine

The force required to stretch the spring $\text{9 cm}$ beyond its natural length.

Answer to Problem 35E

The force required to stretch the spring $\text{9 cm}$ beyond its natural length is $18\text{ N}$ .

Explanation of Solution

Given:

A spring needs a force of $\text{6 N}$ to stretch $\text{3 cm}$ beyond its natural length .

Concept Used:

According to Hooke’s law, the force required to stretch or compress a spring by $\text{x units}$ is constant times $\text{x}$ . The formula used to calculate force is:

  $\text{F = kx}$

Where;

  $\text{F}$ is the force required.

  $\text{k}$ is constant which is called force constant.

  $\text{x}$ is distance by which spring stretch or compress from its natural length.

Calculation:

The force ( F₁ ) is $\text{6 N}$ which is required to stretch the spring by distance $\text{3 cm}$ ( x₁ ) .

Now, calculate the force ( F₂ ) required to stretch the spring by distance $\text{9 cm}$ ( x₂ ).

According to Hooke’s Law:

  $\begin{array}{c}F{1}_{}=k{x}_1\\ k=\frac{F_1}{x_1}\\ \text{ = }\frac{6N}{3\times {10}^{-2}m}\\ \text{ = 200 }\frac{N}{m}\end{array}$

Also,

  $\begin{array}{c}F{2}_{}=k{x}_2\\ F{2}_{}=200\frac{\text{N}}{\text{m}}\times 9\times {10}^{-2}\text{m}\\ \text{ = }1800\times {10}^{-2}\text{N}\\ \text{ =}18\text{ N}\end{array}$

Conclusion:

The force required to stretch the spring $\text{9 cm}$ beyond its natural length is $18\text{ N}$

(b)

To determine

The work done to stretch the spring $\text{9 cm}$ beyond its natural length.

Answer to Problem 35E

The work done to stretch the spring $\text{9 cm}$ beyond its natural length is $0.8\text{ J}$ .

Explanation of Solution

Given:

A spring needs a force of $\text{6 N}$ to stretch $\text{3 cm}$ beyond its natural length .

Concept Used:

According to Hooke’s law, the force required to stretch or compress a spring by $\text{x units}$ is constant times $\text{x}$ . The formula used to calculate force is:

  $\text{F = kx}$

Where;

  $\text{F}$ is the force required.

  $\text{k}$ is constant which is called force constant.

  $\text{x}$ is distance by which spring stretch or compress from its natural length.

Calculation:

The work done can be calculated as:

  $\begin{array}{l}\text{dW=F}x\cdot \text{dx}\\ \text{W}={\displaystyle \underset{0}{\overset{x}{\int }}kx\cdot dx}\\ \text{W}=k\frac{x^2}{2}\end{array}$

Substitute the values in the above formula:

  $\begin{array}{l}\text{W}=\frac{1}{2}\times 200\frac{N}{m}\times 9\times {10}^{-2}m\\ \text{W}=8100\times {10}^{-4}\\ \text{W}=0.8\text{ J}\end{array}$

Conclusion:

The work done to stretch the spring $\text{9 cm}$ beyond its natural length is $0.8\text{ J}$ .