Chapter 7.1, Problem 36E

(a)

To determine

To Find: The work done needs to be determined in compressing the spring the first half inch.

Answer to Problem 36E

The work done in compressing the spring the first half inch is $1250\text{ lb}\cdot \text{in}$ .

Explanation of Solution

Given information: A 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches.

Formula used: The Hooke’s law is used which says that when a spring is compressed or stretched x units from its natural length then the force it takes is a constant times x and in symbols it is as $F=kx$ where k is measured in force units per unit length.

Calculation:

Step 1:

When a 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches, then by Hooke’s law one has as:

  $\begin{array}{c}F=kx\\ 10000=k\cdot 1\\ k=10,000\text{ lb/in}\end{array}$

Step 2:

Now further when spring is compressed first half inch then the force required is evaluated as:

  $\begin{array}{c}F={\displaystyle \underset{0}{\overset{0.5}{\int }}kx}dx\\ F={\displaystyle \underset{0}{\overset{0.5}{\int }}10000xdx}\\ F=10000{\left[\frac{x^2}{2}\right]}_0^{0.5}\\ F=1250\text{ lb}\cdot \text{in}\end{array}$

Hence the work done in compressing the spring to first half inch is $1250\text{ lb}\cdot \text{in}$ .

(b)

To determine

To Find: The work done needs to be determined in compressing the spring the second half inch.

Answer to Problem 36E

The work done in compressing the spring the second half inch is $3750\text{ lb}\cdot \text{in}$ .

Explanation of Solution

Given information: A 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches.

Formula used: The Hooke’s law is used which says that when a spring is compressed or stretched x units from its natural length then the force it takes is a constant times x and in symbols it is as $F=kx$ where k is measured in force units per unit length.

Calculation:

Step 1:

When a 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches, then by Hooke’s law one has as:

  $\begin{array}{c}F=kx\\ 10000=k\cdot 1\\ k=10,000\text{ lb/in}\end{array}$

Step 2:

Now further when spring is compressed second half inch then the force required is evaluated as:

  $\begin{array}{c}F={\displaystyle \underset{0.5}{\overset{1}{\int }}kx}dx\\ F={\displaystyle \underset{0.5}{\overset{1}{\int }}10000xdx}\\ F=10000{\left[\frac{x^2}{2}\right]}_{0.5}^1\\ F=3750\text{ lb}\cdot \text{in}\end{array}$

Hence the work done in compressing the spring to second half inch is $3750\text{ lb}\cdot \text{in}$ .