A stone of 1kg is thrown with a velocity of 20 m/s across the frozen surface of lake and comes to rest after traveling a distance of 50m . what is the force of friction between the stone and the ice???
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Answer. u = 20m/s.
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Answer:
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
We know that
F = m×a
Substituting above obtained value of a = -4 in F = m x a we get,
F = 1 × (-4) = -4N
(Here the negative sign indicates the opposing force which is Friction
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