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05/01/2021 7:22 pm
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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?
1 Answer
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05/01/2021 7:28 pm
Here, m = 1 kg, u = 20 ms-1
v = 0, s = 50 m, F = ?
From, v2 - u2 = 2as
0 - (20)2 = 2a x 50 = 100 a
or a = \(\frac{-400}{100}\) = -4ms-2
Then, force of friction, F = ma
∴ F = 1(-4) = -4 N
Negative sign indicates that force of friction is opposing the motion of the ball.