The initial velocity of the stone, u= 20 m/s
The final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m
The force of friction between the stone and the ice
We know the third equation of motion
v² = u² + 2as
Substituting the known values in the above equation we get,
0² = (20)² + 2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
We know that
F = m×a
Substituting above obtained value of a = -4 in F = m x a we get,
F = 1 × (-4) = -4N
(Here the negative sign indicates the opposing force which is Friction)