Q. 64.4( 164 Votes )

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?

Answer :

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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