A 1020-kg car is being driven up a 9.26 degree hill. The frictional force is dir

Question

A 1020-kg car is being driven up a 9.26 degree hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 504 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 F_N directed perpendicular to the road surface. The length of the road up the hill is 248 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 135 kJ? Number Units

Explanation / Answer

Wnet = WF + Wg + Wf + Wn

Wnet = Fdcos0 + mgdcos(90+) + 504*d*cos180 + Fn*d*cos90

Wnet = Fd + mgdcos(90+) + 504*d*cos180 + 0

135000 = F*248 + 1020*9.8*248*cos(90+9.26)-504*248

Calculating, F= 2657 N

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