m = 200 kg, R = 20 m, Angle of banking =ФSpeed = v =10 m/s.

Let f be the friction force and N be the normal force from the ground. When the banking of road is done for speed v, it means that the normal force supplies the centripetal force. Friction plays no part. If the speed is either less or more than v, then it comes into play.

Centripetal force =N sinФ = m v²/RAlso, N Cos Ф = mg=> TanФ = v²/gR = 10²/(10*20) = 1/2 cosФ = 2/√5 SinФ = 1/√5 N = mg SecФ = 200*10*√5/2 = 1000√5 Newtons

When the vehicle is moving at 5 m/s less than the optimum speed,the centripetal force required is less. But the normal reaction force N is same.Hence, a friction force f acts upwards (towards outside of the curved road) the incline.

So N CosФ + f sinФ = m g => N (CosΦ + μ sinΦ) = m g N sinФ - f cosФ = m u²/R => N (sinΦ -μ CosΦ) = m u²/R=> (CosФ +μ SinΦ) * u²/gR = (sinФ -μ CosΦ)=> μ [sinΦ *u²/gR + cosФ] = [ sinФ - CosФ * u²/gR ]=> μ = [sinΦ - cosΦ * u²/gR] / [sinФ * u²/gR + cosФ]

=> μ = [1/√5 - 2/√5 * 5²/(10*20)] / [1/√5 * 5²/(10*20) + 2/√5] = [ 3/(4√5) ] / [ 17/(8√5) ] = 6/17

Normal force N = mg/[CosΦ +μ SinΦ] = 200*10/ [2/√5 + 6/17 * 1/√5] = 850√5 NFriction force = f = μ N = 300√5 N upwards the slope.