Chapter 8.1, Problem 30E

a. $$

To determine

Tofindthe work done in compressing the spring in the first half inch.

Answer to Problem 30E

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

Explanation of Solution

Given information:

Given a $10,000lb$ force compressed a spring from its natural length of $12inches$ to a length of $11inches$ .

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 xunits from its natural length is given by

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

Given a $10,000lb$ of forces are required to compress a spring $1inch$ from its natural length,

Here $F=10,000;x=1$ then by Hooke’s Law,

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

The work done in compressing the spring in the first half inch is given by,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{0.5}{\int }}10,000xdx}=10,000{\left[\frac{x^2}{2}\right]}_0^{0.5}\\ =1250\text{lb}\cdot \text{in}\end{array}$

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

b. $$

To determine

To find the work done in compressing the spring in the second half inch.

Answer to Problem 30E

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

Explanation of Solution

Given information:

Given a $10,000lb$ force compressed a spring from its natural length of $12\text{inches}$ to a length of $11inches$ .

Formula:

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

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

Given a $10,000lb$ of forces are required to compress a spring $1inch$ from its natural length,

Here $F=10,000;x=1$ then by Hooke’s Law,

  $\begin{array}{l}F=kx\\ 10,000=k\cdot 1\\ \Rightarrow k=10,000lb/in.\end{array}$

The work done in compressing the spring in the second half inch is given by,

  ${\displaystyle \underset{0.5}{\overset{1}{\int }}10,000xdx=10,000{\left[\frac{x^2}{2}\right]}_{0.5}^1}=3750lb.in$

Therefore, the work done in compressing the spring in the first half inch is 3750 lb.in.