Percent error using acceleration and 1/mass

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SUMMARY

The discussion focuses on calculating percent error in a physics lab involving acceleration and mass. The relationship between net force (F), mass (m), and acceleration (a) is established using the formula F = ma, leading to the conclusion that acceleration can be expressed as a = F * (1/m). The slope of the graph plotting acceleration against 1/mass represents the net force, which serves as the accepted value for calculating percent error. The participants confirm that the force applied remains constant across all data points, ensuring accurate results.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with graphing linear equations (y = mx + b)
  • Basic knowledge of percent error calculations
  • Experience with experimental physics involving force and acceleration
NEXT STEPS
  • Study the relationship between force, mass, and acceleration in more complex systems
  • Learn how to derive and interpret slopes from experimental data
  • Explore advanced techniques for minimizing experimental error
  • Investigate the implications of varying mass on acceleration in practical applications
USEFUL FOR

Students in physics labs, educators teaching mechanics, and anyone interested in experimental data analysis and error calculation methods.

mjohnston2
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I'm doing a lab where I had to use a cart and a pully, collecting data to compare the net force and the acceleration and again for acceleration vs mass. The acceleration vs mass graph was a curve, which we then straightened out to give us a graph of acceleration vs 1/mass. I am wondering how i would go about to find my accepted value to use in my percent error equation. thank you ahead of time :)
 
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Welcome to PF, mj.
Since F = ma, a = F*1/m.
Comparing that with good old y = mx + b, the slope on the a vs 1/m graph should be F.
I trust you used the same force for all the data points. The force you used is your "accepted value" for the slope.
 
Delphi51 said:
Welcome to PF, mj.
Since F = ma, a = F*1/m.
Comparing that with good old y = mx + b, the slope on the a vs 1/m graph should be F.
I trust you used the same force for all the data points. The force you used is your "accepted value" for the slope.

YES ! i totally understand :) my slope is in Newtons so it would turn out to be F ! Thank you !
 

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