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A body of mass M starts sliding down on the inclined plane where the critical angle is ÐACB = 30° as shown in figure. The coefficient of kinetic friction will be inclined plane (A) Mg/ (B) Mg (C) 3mg (D) None of these

A body of mass M starts sliding down on the inclined plane where the critical angle is ÐACB = 30° as shown in figure. The coefficient of kinetic friction will be inclined plane

(A) Mg/

(B) Mg

(C) 3mg

(D) None of these

Grade:

2 Answers

Om
11 Points
3 years ago
Frictional force = mgsintheta ( since both are opposite to each other)Umgcostheta = mgsinthetaSintheta / costheta = uU = tantheta Theta = 90-30 = 60U = tan60U = √3I think it will be helpful to you thank
Yokesh K.K.
17 Points
2 years ago
Friction force is given by (mu)mgcostheta
This is equal to mgsintheta
Therefore , mu = square root of 3

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