Answer 1

(a) Normal contact force on the track at B and D is 1000 N.

(b) Force of friction exerted by the track on the cycle at B and D is 25 N.

(c) Normal force between the road and the tyres at C is 1000 N.

(d) Minimum friction coefficient between the road and the tyre is 0.025.

Answer 2

1. Speed Conversion: Convert 18 km/hr to m/s.
2. Normal Force (Horizontal Track): On a horizontal track, the normal force always balances the weight of the object, as there is no vertical acceleration.
3. Friction Force (Centripetal Force): For an object moving in a horizontal circle, the centripetal force required for circular motion is provided by the static friction force. Calculate this using $F_c = Mv^2/R$.
4. Minimum Friction Coefficient: For safe turning without slipping, the required centripetal force must be less than or equal to the maximum static friction force ($f_{s,max} = \mu_s N$). Set them equal to find the minimum coefficient.