A 1300-kg car is being driven up a 4.5° hill. The frictional force is directed o

Question

A 1300-kg car is being driven up a 4.5° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f = 522 N. A force F with arrow 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 with arrow and the normal force F with arrowN directed perpendicular to the road surface. The length of the road up the hill is 270 m. What should be the magnitude of F with arrow, so that the net work done by all the forces acting on the car is +160 kJ?

Explanation / Answer

here

Since, there is no Fn force acting on the car =0J

Fnet = 160 * 10^3 / 270 = 592.6 N

F = Fnet + m * g * sin(4.5) + Fk

F = 592.6 + 1300 * 9.8 * sin(4.5) + 522

F = 2114 J

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