Given, initial velocity of the stone, u=20 ms−1
final velocity of the stone, v=0, mass,
massof stone, m=1kg and
distance covered by the stone, s=50 m.
Let the acceleration be a.
By using third equation of motion, v2−u2 = 2as, we get,
0−202=2a×50
⇒a=−4 ms−2
Since, no external force is acting on a body therefore, force of friction, F=ma=1×–4=−4 N.