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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?
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
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 (finally the stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v2 = u2 + 2as
Where,
Acceleration, a
(0)2 = (20)2 + 2 × a × 50
a = −4 m/s2
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
Hence, the force of friction between the stone and the ice is −4 N.
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 (finally the stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v2 = u2 + 2as
Where,
Acceleration, a
(0)2 = (20)2 + 2 × a × 50
a = −4 m/s2
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
Hence, the force of friction between the stone and the ice is −4 N.
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