A 4.7 kg block slides down an inclined plane that makes an angle of 25 degrees with the horizontal. Starting from rest, the block slides a distance of 2.2 m in 5.2 s. The acceleration of gravity is...

A 4.7 kg block slides down an inclined plane that makes an angle of 25 degrees with the horizontal. Starting from rest, the block slides a distance of 2.2 m in 5.2 s. The acceleration of gravity is 9.81 m/s^2.

Find the coefficient of kinetic friction between the block and plane.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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A 4.7 kg block slides down an inclined plane that makes an angle of 25 degrees with the horizontal. Starting from rest, the block slides a distance of 2.2 m in 5.2 s. The acceleration due to gravity is 9.81 m/s^2. The coefficient of kinetic friction between the block and plane has to be determined.

On the 4.7 kg block, the force of gravity can be divided into two components one parallel to the plane and one perpendicular to the plane.

The perpendicular component is 4.7*9.81*cos 25. If the coefficient kinetic friction is Cf, the force of friction isĀ  4.7*9.81*cos 25*Cf. The parallel component of the gravitational force is 4.7*9.81*sin 25.

The net force acting on the block is 4.7*9.81*sin 25 - 4.7*9.81*cos 25*Cf. The acceleration due to this is 9.81*(sin 25 - cos 25*Cf).

As the block slides down 2.2 m in 5.2 s, 2.2 = (1/2)*9.81*(sin 25 - cos 25*Cf)*(5.2)^2

4.4/5.2^2 = (sin 25 - cos 25*Cf)

=> Cf = 0.2867

The coefficient of kinetic friction between the block and the plane is 0.2867.

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