Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius r designed for speed v0.

"v0" means that there are no friction forces at that speed
mgsinΘ = (mv0²/r)cosΘ → the variable m cancels 
sinΘ/cosΘ = tanΘ = v0² / gr 
Θ = arctan(v0² / gr) 

When v > v0, friction points downslope: 
mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above. 

When v > v0, friction points upslope: 
mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: 
gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ 
µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) 
where Θ is defined above. 

The answer is d) by requesting advice on science internet sites

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x