You are an engineer designing a roller coaster, a section of which is shown on t

Question

You are an engineer designing a roller coaster, a section of which is shown on the right. You want the cars 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 converted 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 (here's a link to the image since it wouldn't let me upload here http://i46.tinypic.com/1zbe59f.jpg

Explanation / Answer

KE + PE = 1/2*m*v^2 + m*g*h = constant (except for the kineticenergy that they tell you is lost) divide by m and you get: .5*v^2 + g*h = constant g = 9.8 m/s^2 v1 = unknown h1 = 6.25m v1= 5.0 m/s h2 = 35.0 m .5 * (5.0 m/s)^2 + 9.8 m/s^2 * 35 m = .5 * v1^2 *(1-.12)+ 9.8 m/s^2*6.25 m simplify simplify 12.5 m^2/s^2 +343 m^2/s^2 = .5 * v2^2 * .88+61.25m^2/s^2 subtract61.25 m^2/s^2 from boths sides 294.25 m^2/s^2 = .44 * v1^2 divideboth sides by .44 668.75m^2/s^2 = v2^2 v1 = ?(668.75319m^2/s^2) v1= 25.86 m/s v1= 25.86 m/s

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