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

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Homework Help Overview

The discussion revolves around a problem related to Newton's laws of motion, specifically focusing on calculating distance using forces and work-energy principles. The original poster is questioning the correctness of their solution, which yields a distance of 3.0 m.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method used to calculate the distance, including the application of forces and work-energy equations. There are concerns raised about the precision of the values used in calculations and the potential for significant error in the results.

Discussion Status

Some participants have confirmed arriving at the same result using a similar method, while others have pointed out issues with the precision of the calculations. There is an acknowledgment of the method's soundness despite the noted concerns.

Contextual Notes

Participants highlight the importance of significant figures in calculations and express confusion regarding the use of the variable W for different purposes within the same context. There is an indication that the original poster may have constraints related to the clarity of their written work versus typed calculations.

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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