When the car is turning the corner the centripetal force is provided solely by friction. If the frictional force was not equal to the centripetal force the car would slip outwards.

The centripetal force if the car of mass m is traveling at a velocity v around a circular path of radius r is m*v^2/r. The frictional force is N*C where N is the normal force m*g and C is the coefficient of friction. Here, the coefficient of friction is 0.4

Substitute the given values in the equation m*v^2/r = m*g*C

=> v^2/5 = g*0.4

=> v^2 = 2*9.8

=> v = 4.427 m/s

The car has to be driven at a speed less than 4.427 m/s if it has to complete the turn without slipping.

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