Chapter 7.5, Problem 8E

To determine

To find: $(a)$ The force constant, $(b)$ How much work is done in stretching the spring, $(c)$ The distance beyond its natural length.

Answer to Problem 8E

  $\begin{array}{c}\text{ a) }200\frac{lb}{in}\\ \text{b) }400\text{ lb-in}\text{. }\\ \text{c) }8\text{ in }\end{array}$

Explanation of Solution

Given information:

The spring has a natural length $10\text{in}$ .

An $\text{800-lb}$ force stretches the spring to $14\text{ in}\text{.}$

Calculation:

For item $(a)$

:

Use Hooke's Law for Springs, which is provided by the formula, to determine the force constant.

  $F=kx\mathrm{.....}(1)$

The natural length $x_0=10\text{ in}$ .

The final length $x_f=14\text{ in}$ due to force

  $F=800\text{lb}$ .

To use Equation $\text{(1)}$ to solve the force constant of the spring,

  $\begin{array}{c}800=k(14-10)\\ \Rightarrow k=\frac{800}{4}\\ \Rightarrow k=200\end{array}$

Since, the force constant of spring is $200\frac{lb}{in}$ .

For item $(b)$

:

 Use the formula to calculate the work done by stretching the spring.

  

To use equation to solve for the work done by stretch the spring from its natural state,

  

Since, the work done by spring is $400\text{lb}-\text{in}\text{.}$

For item $(c)$ :

Use Hooke's Law for Springs, which was defined in Equation, to calculate how much the spring will flex beyond its natural state in response to $1600-\text{lb}$ of force $\text{(1)}$ :

  $\begin{array}{c}1600=200x\\ \Rightarrow x=\frac{1600}{200}\\ \Rightarrow x=8\end{array}$

Since, the spring will stretch by $8\text{ in}$ beyond its natural state due to $1600-\text{lb}$ force.

Therefore, the required $(a)$ the force constant of spring is $200\frac{lb}{in}$ .

  $(b)$ the work done by spring is $400\text{lb}-\text{in}\text{.}$

  $(c)$ the spring will stretch by $8\text{ in}$ beyond its natural state due to $1600-\text{lb}$ force.