Question

What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static friction between tires and road is 0.50?

Answer 1

Answer:

         v= 21.47m/s      

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

$U*m*g=\frac{mv^2}{r}$

$U*g=\frac{v^2}{r}$

substituting our given data in to expression we can solve for the speed V

$0.5*9.81=\frac{v^2}{94}$

making v the subject of formula we have

$0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47$

v= 21.47m/s

hence the maximum velocity of the car is 21.47m/s