http://i80.photobucket.com/albums/j172/XwakeriderX/a.jpg In case you cant read i

Question

http://i80.photobucket.com/albums/j172/XwakeriderX/a.jpg

In case you cant read it

You are an engineer designing a roller coaster, a section of which is shown on the right. You want the cats to be moving at 5.00 m/s at the top of the big hill which is 35.0 m above the ground The bottom of the "dip" just before the big hill is 6.25 meters above the ground. Measurements show that 12.5% of the initial KE of the cars will be convened to heat and sound between the "dip" and the top of the big hill. How fast should the cars be moving at the bottom of the dip? Given: 1 00 m/s = 2.24 MPH

Explanation / Answer

h0 = 6.25 m, h = 35.00 m, v = 5.00 m/s, = 12.5%, find v0

mv02/2 * (1 - ) + mgh0 = mv2/2 + mgh

v02 * (1 - ) + 2gh0 = v2 + 2gh

v0 = [(v2 + 2g(h - h0)/(1 - )] = 25.9 m/s

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