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    Cyclist's speed at crest of hill & frictional force on motion of a car

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    A 82.3-kg skier rides a 2860-m-long lift to the top of a mountain. The lift makes an angle of 16.6° with the horizontal. What is the change in the skier's gravitational potential energy?

    A cyclist approaches the bottom of a gradual hill at a speed of 28.7 m/s. The hill is 8.84 m high, and the cyclist estimates that she is going fast enough to coast up and over it without pedaling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

    A 1100-kg car is being driven up a 9.21° 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 230 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 198 kJ?

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    Solution Preview

    1.
    E=m*g*h=m*g*L*sin(theta)=82.3*9.8*2860*sin(16.6°)=658673.9 (joule)

    2.
    From the convervation of energy, ...

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