Question
Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?
1. $1.23\times 10^{3}\textrm{ N/m}$
2. $6.88 \textrm{ kg}$
3. $4.00 \textrm{ mm}$

OpenStax College Physics, Chapter 16, Problem 1 (Problems & Exercises)

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You have the answer for part (b) as the answer for part (a).

Is it possible to do this problem using the same formula, but taking the magnitude of (F) and (X) and dropping the negative before the (K)?

Yes, the calculation would work out to the same positive result. You can do this, and in Newton's Second Law questions (not Hook's Law questions like this) I usually take the magnitude of the forces and put negative signs in the equation, which is similar to the technique you're proposing of using magnitudes for displacement ($x$ in this equation) and changing the equation to suite (by removing the negative in front of the $k$). For my own way of thinking I would find your proposal confusing since I like the negative in the equation to remind me that the direction of the spring force is opposite to the direction of the displacement of the end of the spring, but if you find a technique that reliably gives you the correct answer and you can explain why it works then go ahead with it.
All the best,
Shaun

Why are we using "g" as 9.81 now when we have been using it as 9.80?

Hi jsotosky, thank you for the question. Sorry for the inconsistency. The textbook often writes 9.80, but 9.81 is more often found elsewhere, such as wikipedia. It actually depends where on Earth you're located. $g$ is slightly greater at the poles, which are closer to the center of the Earth, whereas $g$ is less on the equator where the central bulge of the Earth makes you further from the center.
All the best,
Shaun