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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?
Question by OpenStax is licensed under CC BY 4.0.
  1. $1.23\times 10^{3}\textrm{ N/m}$
  2. $6.88 \textrm{ kg}$
  3. $4.00 \textrm{ mm}$
Solution Video

OpenStax College Physics Solution, Chapter 16, Problem 1 (Problems & Exercises) (2:57)


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Submitted by NoOneReally on Sun, 06/14/2020 - 20:51

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

Submitted by Chelsea Nikkol… on Tue, 11/30/2021 - 08:55

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)?

Submitted by ShaunDychko on Mon, 12/06/2021 - 07:50

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,

In reply to by Chelsea Nikkol…

Submitted by Chelsea Nikkol… on Tue, 11/30/2021 - 14:13

where did the .5 come from in part (c)