How Do You Calculate the Coefficient of Kinetic Friction for a Moving Crate?

AI Thread Summary
To calculate the coefficient of kinetic friction for a crate being pushed with a horizontal force of 210 N, the net force is zero since the crate moves at constant velocity. The friction force equals the applied force, which is 210 N. The equation used is 210 = (muk)(20 kg)(9.8 m/s²), leading to a calculated coefficient of kinetic friction (muk) of 1.07. It is noted that a coefficient greater than 1 is possible on certain surfaces, indicating that the calculations may not be incorrect. The discussion highlights the importance of considering surface characteristics when interpreting friction coefficients.
Mebmt
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Homework Statement


Horizontal force of 210 N used to push 20 kg crate over 4 m with constant velocity. Looking for coefficient of kinetic friction.

Homework Equations


a=0
Work done by friction force is:
Net force=ma
Fapp-Fk=ma
210-Fk=0
Friction force = 210

The Attempt at a Solution



friction force=(muk)mg
210=(muk)(20)(9.8)
muk=1.07

Where did I go wrong?
Scott
 
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Mebmt said:

Homework Statement


Horizontal force of 210 N used to push 20 kg crate over 4 m with constant velocity. Looking for coefficient of kinetic friction.


Homework Equations


a=0
Work done by friction force is:
Net force=ma
Fapp-Fk=ma
210-Fk=0
Friction force = 210

The Attempt at a Solution



friction force=(muk)mg
210=(muk)(20)(9.8)
muk=1.07

Where did I go wrong?
Scott
assuming a horizontal surface, looks OK. muk can be greater than 1 for certain rough or rubbery surfaces.
 

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