Your team of test engineers is to release the parking brake so an 792.0-kg car w

Question

Your team of test engineers is to release the parking brake so an 792.0-kg car will roll down a very long 6.1 percent grade in preparation for a crash test at the bottom of the incline. (On a 6.1 percent grade the change in altitude is 6.1 percent of the horizontal distance traveled.) The total resistive force (air drag plus rolling friction) for this car has been previously established to be F_d = 100 N + (1.2 N Middot s^2.m^2)V^2, where v is the speed of the car. What is the terminal speed for the car rolling down this grade? m/s

Explanation / Answer

At terminal speed, forces are balanced,

mg sin theta = Fd

792*9.8*0.061 = 100 +1.2 v^2

v^2 = [792*9.8*0.061-100]/1.2 = 311.2

v = sqrt(311.2) = 17.64 m/s

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