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Question

A stone of 1 kg is thrown with a velocity of 20 ms1 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?


A

5 N

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B

3 N

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C
There is no friction between stone and ice
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D

4 N

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Solution

The correct option is D

4 N


Initial velocity of the stone = 20 ms1.
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= 4ms2.

The negative sign indicates that acceleration is in the opposite direction to the motion of the stone.
Mass of the stone, m= 1 kg.
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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