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Question
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?
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Solution
Mass of the stone (m) = 1kg
Distance travelled (s) = 50 m
Initial velocity of the stone (u) = 20 m/s
The final velocity of the stone (v) = 0 m/s
From the third equation of motion
v2 - u2 = 2as
02 - (20)2 = 2 × a × 50
02 - (20)2 = 100a
a = `(-400)/100`
a = -4 m s-2
The frictional force between the stone and ice
F = m × a
F = 1 × -4
= -4 N
Therefore, the frictional force between the stone and ice = -4 N
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