Your team of test engineers is to release the parking brake so an792.0-kg car wi

Question

Your team of test engineers is to release the parking brake so an792.0-kg car will roll down a verylong 6.2 percent grade in preparationfor a crash test at the bottom of the incline. (On a 6.2 percent grade the change in altitude is6.2 percent of the horizontal distancetraveled.) The total resistive force (air drag plus rollingfriction) for this car has been previously established to beFd = 100 N + (1.2 N ·s2/m2)v2, wherev is the speed of the car. What is the terminal speed forthe car rolling down this grade?
m/s

Explanation / Answer

Let terminal velocity be v , Also the component of weight along the plane is responsible for thedownward motion of car . component of weight along the plane=mg sin Here tan =0.062 As is very small , so tan = sin= . So component of weight along the plane=mg sin=792.0*9.8*0.062=481.22 N Also at terminal velocity net force is zero . So 481.22 = 100 N + (1.2 N ·s2/m2)v2  1.2v2 = 381.22 So v=17.8 m/s

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