Initial velocity of the stone, u = 20 m/s

Final velocity of the stone, v = 0 m/s

Distance covered by the stone, s = 50 m

According to the third equation of motion:

v ^{2} = u^{ 2} + 2as

Where,

Acceleration, a

(0)^{2} = (20)^{2} + 2 × a × 50

a = – 4 m/s^{2 }

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m = 1 kg

From Newton's second law of motion:

Force, F = Mass x Acceleration

F= ma

F= 1 × (– 4) = – 4 N

Hence, the force of friction between the stone and the ice is – 4 N.