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Question

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

5N

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B

3N

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C
There is no friction between stone and ice
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D

4N

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Solution

The correct option is D

4N


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=20×20+2×a×50

a=4 ms2.

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 = mass × acceleration. Therefore,

F = 1 × (-4) = - 4 N

Hence, the force of friction between the stone and the ice is 4 N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.


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