Beginnger Coefficient of Kinetic Friction Problem

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SUMMARY

The forum discussion centers on calculating the coefficient of kinetic friction (μk) for a box being pushed across a floor. Given a horizontal force of 28N and a box mass of 10.2 kg, the normal force (FN) is determined to be 100N. The kinetic friction force (Ffk) is equal to -28N, leading to a calculated μk of -0.28. The correct value for μk is 0.28, as the coefficient of friction is always expressed as a positive value.

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  • Knowledge of friction forces and coefficients
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Homework Statement



If you exert 28N of horizontal force while pushing a 10.2 kg box across the surface of a floor at constant velocity, then what is the kinetic friction coefficient between floor and the box?

Homework Equations



EF = ma
Ffk = uK * Fn


The Attempt at a Solution



Fg = 10.2 * -9.8 = -100 = 100 = FN
Ef = ma
28N + Ffk = 10.2 * 0 m/s^2 (constant velocity)
Ffk = -28N

uk = -28/100 = -.28 coefficient is this correct?
 
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Perfectly correct, except for the minus sign. μN just gives the magnitude of the friction force; μ is always positive.
 
thanks for the help
 

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