A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s². The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is P/10. The value of P is:

The FBD is shown in the figure. (Note: A figure would be included in a properly formatted solution, showing the forces acting on the ring. These forces would include the applied force from the stick (2N), the frictional force between the stick and the ring (fs), the frictional force between the ground and the ring (fk), and the weight of the ring (mg)).

Moment of inertia I = mR² = 2 * (0.5)² = 0.5 kgm²

N - fs = ma = 2 * 0.3 = 0.6
2 - 0.6 = fs
1.4 = fs

a = Rα = R(τ/I) = R(fs-fk)R/I = (0.5)²(fs-fk)/0.5
fs - fk = 0.3 * 0.5 / (0.5)² = 0.6
1.4 - fk = 0.6
=> fk = 0.8
fk = μN = (P/10) * 2
=> 0.8 = (P/10) * 2
=> P = 4