# A block accelerates at 3.6 m/s^2 down a plane inclined at an angle 24.0 degrees. Find k between the block and the inclined plane. The acceleration of gravity is 9.81 m/s^2

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### 1 Answer

It is give that a block placed on an inclined plane accelerates at 3.6 m/s^2 down the plane. The angle of inclination is 24.0 degrees.

The coefficient of kinetic friction k needs to be determined.

The gravitational force of attraction due to the Earth acting on the block accelerates it vertically downwards at 9.81 m/s^2.

This force has to be divided into its components parallel to the plane and perpendicular to the plane. If the mass of the block is M, the force parallel to the plane acting downwards is M*9.81*sin 24 and the force acting normal to the plane is M*9.81*cos 24*k which is in the opposite direction. The resultant of the two is a force equal to M*3.6. This gives:

M*3.6 = M*9.81*sin 24 - M*9.81*cos 24*k

=> 3.6 = 9.81*sin 24 - 9.81*cos 24*k

=> k = (9.81*sin 24 - 3.6)/9.81*cos 24

=> k = 0.0435

**The coefficient of kinetic friction is 0.0435**

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