Bubble in the letter corresponding to your version of the exam. You have version

Question

Bubble in the letter corresponding to your version of the exam. You have version B. A man pulls a 130-kg object down an inclined plane for a distance of 1.5 m using a rope. The tension in the rope is 700 N, and the pulling force is directed parallel to the incline. The coefficient of friction between the object and the inclined plane is 0.800, and the angle of inclination with the horizontal is theta = 23.0 degree. What is the net work done on the object? 1900J 1050J 410 J 970 J 390 J with a mass of 45 kg and the other with a mass of 60 kg, push off against After they have pushed off, the relative speed at

Explanation / Answer

N = mg cos23 = 1172.72 Hence friction force = 0.8*1172.72 = 938.18 N ;;; net force = 700+ mgsin23 - 938.18 = 259.61 Hence net work done = Fx 259.61*1.5 = 389 so answer is 390J (E)

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