A 1530-kg car is being driven up a 6.84 Chapter 06, Problem 12 A 1530-kg car is

Question

A 1530-kg car is being driven up a 6.84

Chapter 06, Problem 12 A 1530-kg car is being driven up a 6.84 degree hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 544 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force Fn directed perpendicular to the road surface. The length of the road up the hill is 287 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 185 kJ?

Explanation / Answer

W friction = - 544*286   = 155584 N

W gravity = -m g sin theta d = -1530*9.81*sin(6.84)*287 = 513031 N

W normal = 0

W Force = F*287

so

F*287 = -1260*9.81*sin(7.32)*228 - 476*228=185000

F *287 = 185000-(513031+155584)  

F = 1685 N

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