A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Given,
Initial velocity, u = 20 m/s
Final velocity, v = 0 (As the stone stops)
Acceleration, a =?
Distance travelled, s = 50 m
We know that,
v2= u2+2as
(0)2 = (20)2 + 2 × a × 50
0 = 400 + 100 a
100 a = - 400
a = - 
a = - 4 m/s2
As,
Force, F = m × a
F = 1 × (-4)
F = - 4 N
Therefore, the value of force is – 4 N. Here, the negative sign indicates that the stone opposes the motion of stone.
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