Question
A stone of 1 kg is thrown with a velocity of 20 ms−1 on the frozen surface of a lake. The stone comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer: Option D
:
D
Initial velocity of the stone = 20ms−1.
Final velocity of the stone, v=0(finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
v2=u2+2as
0 × 0 = 20 × 20 + 2 × a × 50
⇒a=−4ms−2.
The negative sign indicates that acceleration is in the opposite direction to the motion ofthe stone.
Mass of the stone, m=1kg.
From the newton's second law of motion:
Force = mass × acceleration. Therefore,
F=1×−4=−4N.
Hence, the force of friction between the stone and the ice is 4N. The negative sign indicates that friction oppposes the relative motion.
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:
D
Initial velocity of the stone = 20ms−1.
Final velocity of the stone, v=0(finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
v2=u2+2as
0 × 0 = 20 × 20 + 2 × a × 50
⇒a=−4ms−2.
The negative sign indicates that acceleration is in the opposite direction to the motion ofthe stone.
Mass of the stone, m=1kg.
From the newton's second law of motion:
Force = mass × acceleration. Therefore,
F=1×−4=−4N.
Hence, the force of friction between the stone and the ice is 4N. The negative sign indicates that friction oppposes the relative motion.
Was this answer helpful ?
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