Melissa's question at Yahoo Answers regarding Hooke's Law

[MarkFL]

Here is the question:

Calculus 2 Hooke's Law Help?

A bathroom scale is compressed 1/14 in when a 150-lb person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's Law, answer the following:

a.) What is the scale's force constant?b.) How much does someone who compresses the scale 1/8 in weigh?c.) How much work is done compressing the scale 1/8 in?

Any help would be greatly appreciated! :)

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Melissa,

Hooke's law is given as:

(1) \(\displaystyle F=-kx\)

where $x$ is the displacement of the spring from equilibrium and $k$ is a positive constant called the force constant of the spring. Solving (1) for $k$, we may write:

\(\displaystyle k=-\frac{F}{x}\)

We see that $k$ will have units of force per length, and in the case of this problem, this will be pounds per inch. I will also give the results in pounds per feet.

a) If we consider a compression of the spring to be a negative displacement, then using the fact that a 150 lb. person compresses the spring 1/14 in., we the find:

\(\displaystyle k=\frac{150}{1/14}\,\frac{\text{lb}}{\text{in}}=2100\,\frac{\text{lb}}{\text{in}}=25200\,\frac{\text{lb}}{\text{ft}}\)

b) Using (1), and the result from part a), we find:

\(\displaystyle F=-\left(2100\,\frac{\text{lb}}{\text{in}} \right)\left(-\frac{1}{8}\text{ in} \right)=262.5\text{ lb}\)

c) To compute the work done in compressing the spring 1/8 in, we may use:

\(\displaystyle W=\int_{-s}^0 F_s\,dx=-k\int_{-s}^0 x\,dx=-\frac{k}{2}\left[x^2 \right]_{-s}^0=\frac{k}{2}s^2\)

With \(\displaystyle s=\frac{1}{8}\text{ in}\) we find:

\(\displaystyle W=\frac{2100\,\frac{\text{lb}}{\text{in}}}{2}\left(\frac{1}{8}\text{ in} \right)^2=\frac{525}{32}\text{ in}\cdot\text{lb}=\frac{175}{128}\text{ ft}\cdot\text{lb}\)