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?
Mass of the stone = 1 kg
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance travelled by the stone, s = 50 m
Using the third equation of motion:
v2 = u2 + 2as
Where,
Acceleration, a
(0)2 = (20)2 + 2 × a × 50
a = −4 m/s2
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N.
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Fig. 8.11
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Fig. 8.12
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