Posted by ajo on .
A 1200-kg car is being driven up a 5.0degrees hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f=524N. A force F is applied to the car by the road and propelsthe 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 290m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is +150kJ?
Can anyone please give me some hints to do it? THX A LOT!
If there is a net force acting on the car, in the direction of motion, it is accelerating. Consider the net force along the direction of motion. The product of that force and the distance moved is 150*10^3 J. (They specify that). The normal force Fn does no work.
(F - f - W sin 5.0)*290 = 150*10^3 J
f = 524 N and W = M g = 1200*9.8 N.
Solve for F.