Starting from rest , a 10.0 Kg piece of luggage slides 6.00 m down a rough 30. d

Question

Starting from rest , a 10.0 Kg piece of luggage slides 6.00 m down a rough 30. degrees incline. Then it slides along a horizontal surface for some distance. The coefficient of Kinetic friction between the luggage and the surface is 0.300 for both the incline and the horizontal. Assume the luggage has a smooth transition from the incline to the horizontal. This is a non-conservative system. 1. Draw the free-body diagram for the luggage on the incline and on the horizontal. 2. Calculate the force of friction acting on the suitcase on the incline plane. 3. Calculate the force of friction acting on the suitcase on the horizontal surface. 4. Using the relationship between the Work-non-conservative of the two frictions and mechanical energy, determine how far the luggage will slide on the horizontal.

Explanation / Answer

1st Step => sketch the situation showing all forces that act on block including the two components of block's weight (mg), one that is parallel and one that is perpendicular to the incline. 2nd Step => calculate the work done by the component of weight parallel to plane Wp = mg sin 30 = (10)(9.8)(0.5) = 49N work done by 49 force in moving block6m = (49)(6.0) = 294 J ANS (a) 3rd Step => calculate force of kinetic friction = Kf Kf = Normal force x 0.300 Normal force = Wn = mg cos 30 = (10)(9.8)(0.866) = 84.868 N Kf = (84.868)(0.300) = 25.4 N work done = (25.4)(6.00) = -152.4 J ANS (b) {neg sign means Kf direction is opposite to motion direction} The Normal force is the force of the surface of incline perpendicular to block. As this force is *perpendicular* to the direction of motion, NO WORK is done by this force. ANS (c) ie there is NO component of THIS force in the direction of motion. or another way of saying the same thing is: Work = Force x Displacement x Cos T => but Cos T = 0, because T = 90°

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