Newton's Law Problem (Did i do this correctly?)

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The discussion revolves around verifying the solution to a physics problem involving Newton's laws, where the calculated distance is 3.0 m. The method used includes calculating forces and work done, but there are concerns about precision in the calculations, particularly in the subtraction of two values that differ only slightly. It is noted that using the symbol "W" for different quantities can lead to confusion. The importance of maintaining significant figures in calculations is emphasized, as the precision of the inputs affects the reliability of the result. Overall, while the method is deemed sound, attention to detail in notation and precision is crucial for accuracy.
Lori
Was wondering if i got the answer correctly or got it by coincidence (answer is 3.0 m according to answer key)

1. Homework Statement

Homework Equations


F = ma

The Attempt at a Solution


Ff = μk*N
N = mgcos(20)=18.41 = W

W = fd
W = ΔKE = 0-2 = -2
-2/f =d
f = just the x components since y component balance
f= Fmotion -Ff
Fmotion = mgsin20 = 6.7
Ff=0.4*N = 0.4*18.41=7.364
f = 6.7-7.364 = -0.664

d = -2/f = -2/-0.664 = 3.01 ≈ 3.0 m

So, the distance is 3.0 m
 

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Coincidentally I've got the same result. Using the same method.
 
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Lori said:
f = 6.7-7.364
This line should raise a red flag. You are taking the difference of two numbers that only differ by about 10%, yet the precision of one of them is only given to two sig figs. If the range for that is really 6.65 to 6.75 then the error range in the result is ±10%. You should go back and ensure both inputs to this subtraction have at least three sig figs.
Lori said:
= mgcos(20)=18.41 = W
W = fd
Using W for two different things is most confusing.

Other than that, your method is sound.
 
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haruspex said:
This line should raise a red flag. You are taking the diffeence of two numbers that only differ by about 10%, yet the precision of one of them is only give to two sig figs. If the range for that is really 6.65 to 6.75 then the error range in the result is ±10%. You should go back and ensure both inputs to this subtraction have at least three sig figs.

Using W for two different things is most confusing.

Other than that, your method is sound.
Understood! I guess It's much more easier to be precise when I am writing down in pencil than typing on my computer! But, I see where the confusion is!

Thanks!
 
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