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?

Initial velocity of the stone, u= 20 m/s
Final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m

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 awe get,
F = 1×(-4) = -4N

(Here the negative sign indicates the opposing force which is Friction)

 

 

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