# Question about finding $k$ in Hooke's Law

My textbook (Advanced Engineering Mathematics by Dennis Zill) offers the following explanation of Hooke's Law:

By Hooke's Law, the spring itself exerts a restoring force $F$ opposite to the direction of elongation and proportional to the amount of elongation $s$. Simply stated $F=ks$ where $k$ is a constant of proportionality called the spring constant. The spring is essentially characterized by the number $k$. For example, if a mass weighing $10lb$ stretches a spring $\frac{1}{2}ft$, the $10 = k \frac{1}{2}$ implies $k=20 \frac{lb}{ft}$. Neccesarily then, a mass weighing, say, $8lb$ stretches the same spring only $\frac{1}{2}ft$.

I have two issues with this paragraph.

1. The explanation says that to find $k$, you take the weight of the load as your $F$, and solve. For example in this problem, $F = ks$ becomes $10 = k \frac{1}{2}$. My question is, why is $F$ equal to $10$? $10$ is just the weight - I was under the impression that $F$ was a mass times its acceleration ?
2. At the end of the paragraph, it says that a weight of $8lb$ would stretch the same spring only $\frac{1}{2}ft$, which is the same as a $10lb$ weight. Is that a typo? What would an 8 pound weight really stretch to?
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1, Weight is a force. It's mass * acceleration due to gravity ie "g". It's a lot clearer in metric systems where you don't have the same unit (lb) for mass and weight.

2, Yes that's a typo. Hooke's law says that the stretch of a spring is proportional to the force. Double the force (ie weight) double the extension (stretch)

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So whenever I see "weight", for example "the car weighs 2000 pounds", I can read it as "there is a force of 2000lbs acting on it". Is that right? –  Imray Nov 11 '12 at 20:36
@Imray - we use weight only a gravity force. The car weighs 2000lbs = the car is pulled down onto earth with a force of 2000lbs. –  Martin Beckett Nov 11 '12 at 21:10