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?

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](https://en.wikipedia.org/wiki/Gravitational_acceleration#Comparative_gravities_of_the_Earth,_Sun,_Moon,_and_planets). 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

nvm I figured it out, sorry for bugging you.

I'm glad you figured it out, and no worries!

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

And you don't list part c in the answer set....

Thank you so much for spotting this oversight. I've added part (c) to the question and the answer. Best wishes with your studies! Shaun