A 1280-kg car is being driven up a 9.80 ° hill. The frictional force is directed

Question

A 1280-kg car is being driven up a 9.80 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 511 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 287 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 201 kJ?

Explanation / Answer

Here ,

let the force is F

as the work done by the normal force is zero on the car

net work done = work done by F + work done by friction + work done by gravity

201 *10^3 = F * 287 - 511 * 287 - 1280 * 9.8 * sin(9.8) * 287

solving for F

F = 3346.5 N

the magnitude of applied force F is 3346.5 N

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