Calculating the coefficient of kinetic friction

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction using a pulley system involving a wooden block and a weight. Key variables include a time (T) of 1.28 seconds, a mass of the wooden block at 0.243 kg, and a mass of the weight at 0.110 kg, with a distance (d) of 0.5 meters. The relevant equations include D=vitf+1/2atf^2 and F=ma, with the force of kinetic friction defined as the coefficient of kinetic friction multiplied by the normal force. The participant confirmed their method for solving acceleration and calculating the coefficient, indicating a need for clarification on their approach.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with kinematic equations (D=vitf+1/2atf^2)
  • Knowledge of the concept of normal force in physics
  • Basic principles of friction and coefficients of friction
NEXT STEPS
  • Review the derivation of the coefficient of kinetic friction from experimental data
  • Study the application of kinematic equations in real-world scenarios
  • Explore the relationship between mass, weight, and normal force in friction calculations
  • Investigate common experimental setups for measuring friction coefficients
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in experimental physics and the principles of friction.

Baller123
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Homework Statement


This was a lab. We attached a weight to a pulley and connected it to a wooden block. the wooden block was placed on a wooden board on a table. we dropped the weight and calculated the time it to the block to travel a certain distance.We have the variables
T=1.28 s
Mass of wooden block=.243 kg
Mass of weight on pulley=.110 kg
d=.5 m

Homework Equations


D=vitf+1/2atf^2
F=ma
Force of kinetic friction=coefficient of kinetic friction times normal force

The Attempt at a Solution


I solved for acceleration and used that to solve for the coefficient. I'm sure that's wrong
 
Physics news on Phys.org
Your approach sounds ok to me. Provided you didn't just use F from F=MA.
 

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