The coefficient of kinetic friction (μk) for a moving crate can be calculated using the equation μk = Fk / (mg), where Fk is the friction force and mg is the weight of the crate. In this case, a horizontal force of 210 N is applied to push a 20 kg crate, resulting in a calculated μk of 1.07. This value indicates that the surface may be exceptionally rough or rubbery, as coefficients of friction greater than 1 are possible under certain conditions. The calculations confirm that the applied force equals the friction force when the crate moves at a constant velocity.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics, as well as engineers and professionals involved in material science and surface interactions.
assuming a horizontal surface, looks OK. muk can be greater than 1 for certain rough or rubbery surfaces.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