Question

What should be the coefficient of friction between the tyres and the road when car travelling at 60 km/hr and makes a level turn of radius 40 m

Answer 1

Answer:0.71

Explanation:

Given,

Speed of car=60 km/h=60000/3600 m/s=16.7 m/s

Radius of turn=40 m

The maximum centripetal force that friction can provide here, isF=μsN=μs W=μs mgThus,Centripetal force=Frictional forcemv2r=μs mgor μs=v2rg=16.7240×9.8=278.89392=0.71

Answer 2

GIVEN,

speed of car= 60kmph, radius of turn= 40m.

TO FIND,

the coeffecient of friction between tyre and the road.

SOLUTION,

given,

speed= 60kmph

         =  $\frac{60000}{3600}$ m/s ( changing into m/s since we are given radius in m)

         = 16.7 m/s

radius of turn for the car= 40m

here the centripetal force will come in action,

F= μsN

 = μs

W= μs Mg. (work done)

hence,

centripetal force=  μs MgoR

                          μs= V²Rg

                             = 16.7240 * 9.8

                             = 278.89392

                             = 0.71.

HENCE THE  COEFFICIENT OF FRICTION IS 0.71.