A. Suppose thecoefficient of static friction between the road and the tiers on a

Question

A. Suppose thecoefficient of static friction between the road and the tiers on acar is 0.60 and the car hasno negative lift. What speed will put the car on the verge of sliding as itrounds a level curve of 30.5mradius? --------------------------------------------------------------------------------------------------------- What is the velocity of the block when it reached the bottomof the plane. (Hint: Choose your own x-y coordinate system) --------------------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------------------

Explanation / Answer

A.. When the car is on the curve, centripetal force =mv2/r On the Verge of sliding ,this force should be equal to frictionalforce = mg Thus, mv2/r = mg =>v=(gr)       =179.52 m/s       = 13.398 m/s B. Angle of inclination =45 degrees So, vertical height=s= 2 sin 45= 1.732 m Now, Since initial velocity=0, v2=2gs =>v=2*9.8*1.732       = 5.826 m/s

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