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06/01/2021 5:30 pm
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A stone of 1 kg is thrown with a velocity of 20 ms-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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06/01/2021 5:34 pm
m = 1 kg
u = 20 m/s
s = 50 m
v = 0
F = ?
a = ?
\(v^2 - u^2\) = 2as
\((0)^2 - (20)^2\) = 2a(50)
∴ - 400 = 100 a
∴ a = \(\frac{-400}{100}\) = -4 m/s2
Force of friction, F = m x a
= 1 kg x - 4 m/s2
= -4 N