Confused about significant figures in official solutions

In summary, the book states that multiplication or division may have no more significant figures than the starting number with the fewest significant figures. However, when one reports experimental results, they need to track the number of significant figures in their results. This is illustrated with the example of inputs with leading digits of 8 or 9 and outputs with a leading digit of 1.
  • #1
Argonaut
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Homework Statement
[Sears and Zemansky's University Physics, 13E] 6.3 A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?
Relevant Equations
Force and Work
I keep running into the issue of getting a different number of significant figures in my solutions than the official answers at the back of the book, so I'm wondering if I'm missing something or if the book is occasionally sloppy about significant figures.

In Unit 1, the book states (p.8.):
Multiplication or Division:
Results may have no more significant figures than the starting number with the fewest significant figures.

And the these are the answers official answers to the exercise above:
6.3
a) 74 N
b) 333 J
c) -330 J
d) 0, 0
e) 0

Here is my final calculation of a):
$$
F_p = \mu_k m g = (0.25)(30.0~\rm{kg})(9.80~\rm{m/s^2}) = 73.5~\rm{N}
$$
As far as I can tell, each term has three significant figures, unless ##30.0## counts as two significant figures.

And then one of b) or c) has to be a typo as the ##W_p = -W_f##. This is my calculation of ##W_p##:

$$
W_p = F_p s\cos{\phi} = (73.5 ~\rm{N})(4.5 ~\rm{m})(\cos{(0^{\circ})})=330 ~\rm{J}~\text{(to 2 significant places)}
$$
Since the displacement is given to two significant places, ##333 ~\rm{N}## is spurious accuracy, isn't it?
 
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  • #2
I agree with you, it is spurious accuracy. However there is a more serious shortcoming in the answers as written. The crate is moving at constant velocity. According to the work-energy theorem. this means that the work done by the pushing force and the work done by friction must add to zero. According to the solution, they do not. That's not the proper way to teach students through answers to questions.
 
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  • #3
kuruman said:
I agree with you, it is spurious accuracy. However there is a more important issue with the answers as written. The crate is moving at constant velocity. According to the work-energy theorem. this means that the work done by the pushing force and the work done by friction must add to zero. According to the solution, they do not. That's not the proper way to teach students through answers to questions.
Thanks for putting my mind at ease.

What do you think about a)? Is ##73.5~\rm{N}## or ##74~\rm{N}## correct with regards to significant figures?
 
  • #4
Argonaut said:
Thanks for putting my mind at ease.

What do you think about a)? Is ##73.5~\rm{N}## or ##74~\rm{N}## correct with regards to significant figures?
Don't mind me.

I've just realised the coefficient of kinetic friction ##\mu_k = 0.25## is given to two significant places, so ## 74~\rm{N}## is correct after all.

That still leaves the work values as problematic.

Thanks for the help again!
 
  • #5
Argonaut said:
What do you think about a)? Is ##73.5~\rm{N}## or ##74~\rm{N}## correct with regards to significant figures?
I think that 4.50 m is correct for the displacement distance. Then it would be clear how many sig figs one should use for the answers. In this case, 74 N would be strictly correct but only the most nitpicking pedant would mark 73.5 N incorrect.
Argonaut said:
I've just realised the coefficient of kinetic friction ##μ_k=0.25## is given to two significant places, so 74 \rnN is correct after all.
What if you used ##mu_k=\frac{1}{4}##? Also, the value of ##g## was not given. How do you you know how many sig figs to pick for that?

My intention for saying all this is to convey the idea that if you provide one more significant figure than is appropriate in homework problems, it wouldn't matter to most people evaluating your work. It does matter, however, when you report experimental results in which case you need to pay attention to the uncertainty of your measurements and the resultant propagated error.
 
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  • #6
Argonaut said:
Since the displacement is given to two significant places, ##333 ~\rm{N}## is spurious accuracy, isn't it?
The 333 N is the result of using a rounded intermediate result (74) in a subsequent calculation. That is poor practice. Intermediate results should be used in subsequent calculations at full accuracy. [One should still track the number of reliable significant figures in the intermediate result so that the number of reliable significant figures in the final result can be correctly determined].

Using a rounded-to-two-figures intermediate result to compute a final result that is quoted to three figures is just crazy.

The authors of Question b might point to the wording: "how much work is done [...] by this force". I claim that "this force" properly refers to 73.5 plus or minus 0.5 (or a more accurately computed error bar), not 74 plus or minus 0.5 and certainly not to 74.0 plus or minus zero.

Edit: I am completely on board with what @kuruman has to say about keeping an additional significant digit when appropriate. The value for ##g## is a somewhat useful example. The error bound is around the plus or minus 0.25% level (extremes found on the surface of the earth) That is ##-log_{10}\frac{1}{0.025} = 2.6## significant digits. So if someone wants to do ##g=9.81##, that would be fine in my book. Another example is inputs with leading digits of 8 or 9 and outputs with a leading digit of 1 or 2. Adhering to the rules of significant digits in such situations tends to under-report the significance of the result. (8 + 9 = 17 which rounds to 20 -- the result is off by three on average, even though the inputs were only off by 0.5 each at most).
 
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  • #7
Thanks for the insight, both!
 

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