We have the next information
r=radius=142m
k=coeffcient of friction=1.07
The maximum frictional force is

where F is the frictional force, k is the coefficient of friction, m is the mass and g is the gravity.
Also for the centripetal force we have the next formula

where F is the force, m is the mass , v the speed and r is the radius
we have an equivalence between the two formulas

then we simplify and we isolate the speed
![v=\sqrt[]{\text{krg}}](https://img.qammunity.org/2023/formulas/physics/high-school/o0ua8u2pbh4ombo2sppkircsk883oe29r6.png)
we substitute the values
![v=\sqrt[]{(1.07)(142)(9.8)}](https://img.qammunity.org/2023/formulas/physics/high-school/67tzoll3sooo80uz83itovw5qj2z4nxxvx.png)
the maximum speed is

ANSWER
v=38.6 m/s