Posted by billy on .
A 1310-kg car is being driven up a 9.12 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 541 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 337 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 155 kJ?
Wc = m * g = 1310kg * 9.8N/kg = 12838 N.
Fp = 12338*sin9.12o = 2035 N. = Force
parallel to road.
Fv = 12838*cos9.12 = 12676 N. = Force
perpendicular to road. = The normal.
Work = Fn * D = 155,00 J.
Fn * 337 = 155000
Fn = 460 N. = Net force.
Fn = F-Fp-Fk = 460 N.
F-2035-541 = 460
F = 460+2035+541 = 3036 N.