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Consider a person on a sled sliding down a 100 m long hill on a 30° incline. The mass is 20 kg, and the person has a velocity of 2 m/s down the hill when they're at the top. How fast is the person traveling at the bottom of the hill? All we have to worry about is the kinetic energy and the gravitational potential energy; when we add these up at the top and bottom they should be the same, because mechanical energy is being conserved.
in Grade 12 Physical Sciences by Diamond (49,550 points) | 29 views

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At the top:

PEg = mgh

        = (20)(9.8)(100Sin30°) -> (its a 100m long hill, we want the height, which is Sin30 = opp/hyp)

        = 9800J

KE = 1/2mv²

      = 1/2(20)(2)²

      = 40J

ME = PEg + KE

      = 9800 + 40

      = 9840 J

At the bottom:

PEg = 0

KE = 1/2mv²

ME = 0 + 1/2mv²

9840 = 1/2(20)v²

20v² = 19680

v = √19680/20

   = 31.37 m/s
by Wooden (786 points)
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