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# Work, Energy, and Power Name

 tarix 25.06.2016 ölçüsü 35 Kb.
 Work, Energy, and Power Name: A 4768-kg roller coaster train full of riders approaches the loading dock at a speed of 17.1 m/s. It is abruptly decelerated to a speed of 2.2 m/s over a distance of 13.6 m. Determine the retarding force that acts upon the roller coaster cars. KEi + Wext = KEf ½•(4678 kg)•(17.1 m/s)2 + F•(13.6 m)•cos(180°) = ½•(4678 kg)•(2.2 m/s)2 683946.9 … J - 11320.7 … J = F•(13.6 m) F = 49500 N (49457.8 … N) A catcher's mitt recoils a distance of 12.9 cm in bringing a 142-gram baseball to a stop. If the applied force is 588 N, then what was the speed of the baseball at the moment of contact with the catcher's mitt? KEi + Wext = 0 J ½•(0.142 kg)•vi2 + (588 N)•(0.129 m)•cos(180°) = 0 J (0.071 kg)•vi2 = 75.852 J vi = 32.7 m/s An unknown force is applied to a 12 kg mass. The force acts at an angle of 30 degrees above the horizontal. Determine the force acting if the force acts for a horizontal displacement of 22 meters and increases the 12 kg mass's speed from 11 m/s to 26 m/s. KEi + Wext = KEf ½•(12 kg)•(11 m/s)2 + F•(22 m)•cos(0°) = ½•(12 kg)•(26 m/s)2 F•(22 m) = 4056 J - 726 J F= 150 N (151.36 … N) A physics teacher exerts a force upon a 3.29-kg pile of snow to both lift it and set it into motion. The snow leaves the shovel with a speed of 2.94 m/s at a height of 0.562 m. Determine the work done upon the pile of snow. Wext = KEf + PEf W = ½•(3.29 kg)•(2.94 m/s)2 + (3.29 kg)•(9.8 N/kg)•(0.562 m) W = 14.218 … J + 18.120 … J W = 32.3 J A 250.-gram cart starts from rest and rolls down an inclined plane from a height of 0.541 m. Determine its speed at a height of 0.127 m above the bottom of the incline. PEi = KEf + PEf (0.250 kg)•(9.8 N/kg)•(0.541 m) = ½•(0.250 kg)•vf2 + (0.250 kg)•(9.8 N/kg)•(0.127 m) 1.325 … J - 0.311 … J = ½•(0.250 kg)•vf2 vf = 2.85 m/s A 4357-kg roller coaster car starts from rest at the top of a 36.5-m high track. Determine the speed of the car at the top of a loop that is 10.8 m high. KPEi = KEf + PEf (4357 kg)•(9.8 N/kg)•(36.5 m) = ½•(4357 kg)•vf2 + (4357 kg)•(9.8 N/kg)•(10.8 m) 1558498.9 … J - 461144.9 … J = ½•(4357 kg)•vf2 vf = 22.4 m/s © The Physics Classroom, 2009 Page

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