Suppose the roller coaster below (h1=35m, h3=26m, h4=15m) passes point 1 with a

Question

Suppose the roller coaster below (h1=35m, h3=26m, h4=15m) passes point 1 with a speed of 18.0 m/s. if the average force of friction is equal to one seventh of its weight, with what speed will it reach point 4? the distance traveled from 1 to 44 is 150m.

Suppose the roller coaster below (h1=35m, h3=26m, h4=15m) passes point 1 with a speed of 18.0 m/s. if the average force of friction is equal to one seventh of its weight, with what speed will it reach point 4? the distance traveled from 1 to 44 is 150m.

Explanation / Answer

friction force = weigth / 7 = m * 9.8 / 7 = 1.4 * m

so work done by friction = friction * distacne = 1.4 m * 150 = 210m J

energy at A = mgh1 + 0.5 * m * v1^2 = m*( 9.8 * 35 + 0.5*18^2 ) = 505 m J

Let the velocity at D be v ...

so..

energy at C = mgh4 + 0.5 * m * v^2 = m*(9.8*15 + 0.5*v^2 ) = m*(147 + 0.5 v^2 )

now... energy at A - energy at C = work by friciton

505 m - m*(147 + 0.5 v^2 ) = 210m

so.. 0.5 * v^2 = 148

so.... v = 17.20465 m/sec

so.. velocity at D = 17.20465 m/sec

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