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Unformatted text preview: Problems of Work and kinetic Energy 1. A raindrop (m = 3.35 105 Kg) falls vertically at constant speed under the influence of gravity and air resistance. After the drop has fallen 100m, what is the work done a. by gravity and b. by air resistance? Solution: by the raindrop fall down with the constant velocity so (1) but = m + (2) according (1) and (2) we get m + = 0 or mg = f a. work done by gravity: = m.g.h.cos But direction of gravity and motion are the same so = m.g.h.cos = 3.35 105 10 100 = 3.35 102 J b. work done by air resistance = m.g.h.cos But direction of air resistance and motion are different so = m.g.h.cos = 3.35 105 10 100 = 3.35 102 J 2. A sledge loaded with bricks has a total mass of 18kf and is pulled at constant speed by a rope. The rope is inclined at 20 above the horizontal, and the sledge moves a distance of 20m on a horizontal surface. The coefficient of kinetic friction between the sledge and the surface is 0.5 a. What is the tension of the rope? b. How much work is done on the sledge by the rope? c. What is the energy lost due to friction? prepared by duong chhom morokot Page 1 Thus work done by gravity is = 3.35 102 J Thus work done by gravity is =  3.35 102 J Solution: a. tension of rope: The block has constant speed so which and In the vertical direction: + + = 0 or N = P T y (1) In the horizontal direction: + = 0 or T x = f (2) By T x = T. cos = T. cos20 T y = T. sin = T. sin20 P = mg And f = From (1) we get f = (3) Let (3) in (2) we get T.cos20 = T 0.93 = 0.5 ( 18 10 T 0.34) 0.93T = 90 0.17T or 1.1T = 90 so T = = 81.81N b. Work done by the rope W r = T d = 81.81 20 = 1636.36J c. Energy lose due to friction W f = f d But f = T cos 20 = 81.81 0.93 = 76.08N So W f = 76.08 20 = 1521.6J prepared by duong chhom morokot Page 2 Thus the tension rope is T = 81.81N Thus the work done by rope is W r = 1636.36J Thus the Energy lose due to the friction is W f = 1521.6J 3. A block of mass 2.50kg is pushed 2.20m along a frictionless horizontal table by a constant 16N force directed 25 below the horizontal. Determine the work done by a. the applied force, b. the normal force exerted by the table, and c. the force of gravity. d. Determine the total work done on the block. Solution: a. Determine work done by the applied force W = F d cos = 16 2.20 cos25 = 31.9J b. Determine the work done force exerted by the table W N = N d cos but the direction of d and N make an angle of 90 so we get W N = 0 c. Determine the work done by the force of gravity W g = m g d cos but the direction of g and d make an angle of 270 so we also get W g = 0 d. Determine the total work done on the block W = W + W N + W g = 36.9 + 0 + 0 = 36.9J prepared by duong chhom morokot Page 3 Thus the work done by the applied force is W = 31.9J Thus the work done by force exerted is W N = 0J Thus the work done by the force of gravity is W g = 0J Thus the total work done on the block is W = 36.9J 4.4....
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This note was uploaded on 06/01/2011 for the course CIVIL ENGI 101 taught by Professor Sunithay during the Spring '11 term at CSU Northridge.
 Spring '11
 sunithay

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