A 1150-kg car is being driven up a 7.28 ° hill. The frictional force is directed
ID: 1415186 • Letter: A
Question
A 1150-kg car is being driven up a 7.28 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 510 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 213 kJ?
Explanation / Answer
force of gravity
Fg = m*g*sin 7.28 deg
Fg = 1150*9.81*sin 7.28 deg = 1429.57 N
Force of friction
Ff = uk*N = uk*m*g*cos 7.28 deg
Ff = 510 N
work by normal force = 0
W = Ft*s = 213 kJ
213*10^3 = (F - Ff - Fg - Fn)*s
213*10^3 = (F - 510 - 1429.57)*337
F = 510 + 1429.57 + 213000/337
F = 2571.61 N
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