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

Asked by Vishal kumar | 2 years ago |  265

1 Answer

Solution :-

Given, Mass of the stone (m) = 1 kg

Initial velocity (u) = 20 m/s

Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)

Distance travelled by the stone (s) = 50 m

As per the third motion of equation, (v2 - u2) = 2as

The acceleration of the stone is given by :  \( \frac{(v^2 - u^2)}{2s}\)

Therefore, acceleration of the stone (a) = \( \frac{0-400}{100}ms^2 =-4 \,ms^{-2}\)

As per the second law of motion, F = ma

Therefore, force acting on the stone, \( F = 1kg\, \times\,-4ms^2=-4\,N\)

The frictional force acting on the stone has a magnitude of 4 N and it acts on the direction opposite to that of stone's motion

Answered by Shivani Kumari | 2 years ago

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