Question

Consider a wet banked racetrack, where there is a coefficient of static friction of 0.33 and a coefficient of kinetic friction of 0.28 between the tires and the surface of the racetrack. the radius of the curve is 60.0 m .

Answer 1

Question is missing, but I think the problem is asking for the maximum speed the car can go without going off the road.

The force responsible of keeping the car in the turn is the frictional force, which points towards the center of the turn:

`F_f = \mu m g`
where
`\mu` is the coefficient of static friction, because we require the car not to move in the direction center of the turn-car. In order for the car to stay in the turn, this frictional force must be larger than the centripetal force, given by

`F_c = (mv^2)/(r)`
Therefore, we require

`F_f \ \textgreater \ F_c`

`\mu m g \ \textgreater \ m (v^2)/(r)`
that becomes

`v\ < \ √(\mu g r)`
where r is the radius of the turn. Using
`\mu=0.33` and
`r=60.0m` we find

`v< √(\mu g r)= √(0.33 \cdot 9.81m/s^2 \cdot 60.0m)=13.9m/s`