A 1500-kg car is being driven up a 7.43 ° hill. The frictional force is directed

Question

A 1500-kg car is being driven up a 7.43 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 503 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 326 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 167 kJ

Explanation / Answer

net work done= 167 *10^3 J
so net force = 167000/326 = 512.27 N
also net force = F -503 - 1500*9.8*sin7.43 = F - 2403.93N = 512.27
so, F= 2916.2N

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