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A stone of 1 kg is thrown with a velocity of 20m s-¹ across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice ?
Most Upvoted Answer
A stone of 1 kg is thrown with a velocity of 20m s-¹ across the frozen...
M = 1kg
u = 20m/s. v = 0m/s.
s(distance travelled) = 50m
using third equation of motion
v²=u²+2as
0² = (20)²+2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
F = m×a
F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)....
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A stone of 1 kg is thrown with a velocity of 20m s-¹ across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice ?
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