Question
Tue September 01, 2009 By: Sukhmaneet Kaur

Please solve the numerical

Expert Reply
Tue September 01, 2009

The component of weight parallel to  the inclined plane, W = mgsin30 = mg/2

The component of weight perpendicular to the inclined plane, i.e. normal reaction, W = mgcos30 = 3mg/2

The kinetic friction force = μW 

F = ma

W - μW = ma

mg/2 - μ3mg/2 = mg/4

1/2 - μ3/2 = 1/4

 μ = 1/23 = 0.28

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