A 1400-kg car is being driven up a 8.95 ° hill. The frictionalforce is directed

Question

A 1400-kg car is being driven up a 8.95 ° hill. The frictionalforce is directed opposite to the motion of the car and has amagnitude of 488 N. A force F is appliedto the car by the road and propels the car forward. In addition tothese two forces, two other forces act on the car: itsweight W and the normalforce FN directedperpendicular to the road surface. The length of the road up thehill is 316 m. What should be the magnitudeof F, so that the net work done by all theforces acting on the car is 137 kJ?

Explanation / Answer

From Newton's laws of motion Fnet = F - mgsin - fr given workdone W = Fnet *s = (F - mgsin -fr ) *s then F - mgsin - fr = W/s = 137*10 3 J/316m= 433.5 N therefore F = mgsin + fr + 433.5                  = (1400)(9.8) sin 8.95 + 488N + 433.5N                 = 3055.5N

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