What is the maximum speed with which a 1200-kg car can round a turn of radius 93.0 m on a flat road if the coefficient of static friction between tires and road is 0.60

The maximum speed is 23.39m/s

Explanation:

Centripetal force Fc is the force acting on a body moving in a circular path and directed towards the centre of the path.

If F = ma

centripetal acceleration a = v²/r

Centripetal force Fc = mv²/r ... (1)

where;

m is the mass of the object

r is the radius.

Since there is friction between the tyre and the road, then the frictional force Ff acts between the surface and this frictional force is the one that tends to oppose the moving force (centripetal force)

Ff = μsR where

μs is the coefficient of static friction

R is the normal reaction which is also equivalent to the weight of the car i.e R = W = mg

Ff = μsmg ... (2)

For the body to be static, the centripetal force must be equal to the frictional force i.e Fc = Ff

mv²/r = μsmg

Making v the subject of the formula;

v²/r = μsg

v² = μsgr

v = √μsgr

Given the following data;

μs = 0.6

g = 9.81m/s²

r = 93.0m

v = √0.6×9.81×93

v = √547.398

v = 23.39m/s

he . if i may ask what i your question exactly?

explanation: