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in Physics by (61.1k points)

A stone of 1 kg is thrown with a velocity of 20ms-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?

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Mass of the stone m = 1 kg.
initial velocity u = 20 ms-1
Final velocity v = 0 (Therefore the stone comes to rest distance travelled S = 50 m)
From third equation of motion
v2 = u2 + 2as
(0)2 = (20)2 + 2a(50)
100 a = -400
∴ a = -4ms-2
Here negative sign shows that there is relation in the motion of stone.
Force of friction between stone and ice = Force required to stop the stone
= ma
= 1 × -4 = -4N or 4N.

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