A 10.0 kg block is on an incline plane of variable angle. Coefficient of Kinetic

Question

A 10.0 kg block is on an incline plane of variable angle. Coefficient of Kinetic Friction is .22. Theta = 34.4

what is the acceleration of the block down the plane if it released from rest?

Suppose the block is given an initial velocity up the incline plane. What is the acceleration of the block as it moves up the plane?

Suppose the block is given an initial velocity up the incline plane when the angle of the plane is half of theta. What is the force of friction and acceleration of the block as it moves up the plane?

Explanation / Answer

Acceleration of block = gsin(theta) - u.g cos(theta) = g (sin34.4 - 0.22*cos34.4) = 0.383 g = 3.75 m/s2

If the block is given an initial velocity up the incline plane, the gravity and friction, both forces will act down.

Acceleration of block = gsin(theta) + u.g cos(theta) = g (sin34.4 + 0.22*cos34.4) = 0.746g = 7.31 m/s2

But since the acceleration acts opposite to velocity direction we take it negative. hence acc = - 7.31 m/s2

Now theta = 17.2, Friction = u.M g cos(theta) = 0.22*10*9.8* cos17.2 = 20.59 N

Accelration = ug cos(theta) +  g sin(theta) = 0.22* 9.8 *cos 17.2 + 9.8 * sin 17.2 = - 4.96 m/s2

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