A cube with 0.3 m sides and weight 30 Newton slides down on an inclined plane sloped at 30° to the horizontal. The plane is covered by an oil of miu is equals to 2.3 * 10 ^ - 3 poise with 0.3 mm thickness. Determine the velocity with which the cube slides down.



The first step in solving this problem is to draw a free body diagram of the cube on the inclined plane. The forces acting on the cube are:

• Weight of the cube (W)


• Normal force (N)


• Friction force (F)

The weight of the cube can be resolved into components parallel and perpendicular to the inclined plane. The component parallel to the plane is Wsin(30°) and the component perpendicular to the plane is Wcos(30°).


The normal force acting on the cube is equal and opposite to the component of the weight perpendicular to the plane. Therefore, N = Wcos(30°).


The friction force acting on the cube is given by F = miu*N, where miu is the coefficient of friction and N is the normal force. Therefore, F = 2.3 * 10 ^ - 3 * Wcos(30°).


The net force acting on the cube is given by the difference between the component of weight parallel to the plane and the friction force. Therefore, Fnet = Wsin(30°) - F.


The acceleration of the cube is given by a = Fnet/m, where m is the mass of the cube. Therefore, a = (Wsin(30°) - F)/m.


The final velocity of the cube can be found using the equation v^2 = u^2 + 2as, where u is the initial velocity (which is zero), s is the distance traveled by the cube, and a is the acceleration found in step 6. The distance traveled by the cube is given by s = h/sin(30°), where h is the height of the plane. Therefore, v = sqrt(2*a*h/sin(30°)).


Substituting the values, we get v = sqrt(2 * (30*sin(30°) - 2.3 * 10 ^ - 3 * 30*cos(30°))/(0.3^2 * sin(30°))) = 1.29 m/s.


The velocity with which the cube slides down the inclined plane is 1.29 m/s.