Chapter 7.5, Problem 10E

To determine

To find: $(a)$ The force of work done by the spring who compresses the scale $1/8\text{ in}\text{.}$ weight , $(b)$ The work done by the spring who compresses the scale $1/8\text{ in}\text{.}$ weight.

Answer to Problem 10E

  $\begin{array}{c}\text{a) }F=300\text{lb }\\ \text{b) }W=18.75\text{lb}-\text{in}\text{.}\end{array}$

Explanation of Solution

Given information:

The bathroom scale is compressed $1/16\text{ in}\text{.}$ when a person $150-\text{lb}$ stand on it.

Calculation:

For item $(a)$

:

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

  $F=kx\mathrm{.....}(1)$

To use the given $F=150\text{lb}$ and

  $x=\frac{1}{16}$ into equation $\text{(1)}$ to solve the force constant of the spring,

  $\begin{array}{c}150=k\left(\frac{1}{16}\right)\\ \Rightarrow k=\frac{150}{\left(\frac{1}{16}\right)}\\ \Rightarrow k=2400\end{array}$

Since, the spring constant of spring is $2400\text{ in-lb}$ .

  $\begin{array}{c}F=(2400)\left(\frac{1}{8}\right)\\ \Rightarrow F=300\end{array}$

Hence, the force needed is $F=300\text{lb}$ .

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,

  

Hence, the work done by the spring is $18.75\text{lb}-\text{in}$ .

Therefore, the required $(a)$ The force of work done by the spring who compresses the scale $1/8\text{ in}\text{.}$ weight is $F=300\text{lb}$ , $(b)$ The work done by the spring who compresses the scale $1/8\text{ in}\text{.}$ weight is $18.75\text{lb}-\text{in}$ .