Assume you are on a luge toboggan that has a regulation mass of 22kg and no brakes. The luge relies partly on friction to slow it down. If the coefficient of kinetic friction between the luge and the horizontal icy surface is 0.012, what is the kinetic friction acting on the luge?
So, what I did is I found the normal force first, then substituted its value in Fkinetic= coefficient of friction X Fnormal and my final answer was 2.6N
Is this correct?
So, what I did is I found the normal force first, then substituted its value in Fkinetic= coefficient of friction X Fnormal and my final answer was 2.6N
Is this correct?
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Ff = µ*Fn = .012*22*9.8 = 2.6 N
Attagirl!
Attagirl!