A car of m = 2000 kg safely take a circular turn of radius 60 m at 22 m/s. The m

Question

A car of m = 2000 kg safely take a circular turn of radius 60 m at 22 m/s. The maximum static friction force acts on the car at this moment. (a) Find the centripetal force acting on the car (b) What is the centripetal acceleration of the car? (c) What is the static coefficient of friction? (d) If suddenly a long wet patch in the road decreases the maximum static friction force to one third of what it had earlier, find the maximum speed the car can travel now (the radius remain the same as above).

Explanation / Answer

a) centripetal force = mv^2 / r = 2000*22^2 / 60

= 16133.33 N

b) acceleration = v^2 / r = 22^2 / 6 0 = 8.067 m/s2

c) centripetal force = friction force

mv^2 / r = umg

u*2000*9.8 = 16133.33

u = 0.82

d) if u = 0.82/3 = 0.274

maximum speed now = sqrt(urg)

= sqrt(0.274*60*9.8) = 12.7 m/s

maximum speed now = 12.7 m/s

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