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posted by  aidansmith on 11/6/2009 11:40:24 AM  |  status: Live  |  Earned Karma: 0

magnitude

Course Textbook Chapter Problem Needs by
General Physics N/A N/A N/A 11/8/2009 at 10:00:00 AM
Question Details:
A 1400-kg car is being driven up a 8.95 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 488 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 316 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 137 kJ?

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posted by JamesD on 11/7/2009 12:24:14 PM  |  status: Live
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Response Details:

 From Newton's laws of motion
  Fnet = F - mgsinθ - fr
given
  workdone  W = Fnet *s = (F - mgsinθ - fr ) *s
then F - mgsinθ - fr  = W/s = 137*10 3 J/316m = 433.5 N
therefore  F = mgsin θ + fr + 433.5
                  = (1400)(9.8) sin 8.95 + 488N + 433.5N
                 = 3055.5N
I hope this helps! Best of luck with the rest of your coursework.
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