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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 ? |
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Answer» Solution :Initial VELOCITY of the stone , u=20 m/s Final velocity of the stone v=o Distance COVERED by the stone S=50 m Since `V^2-u^2` 2= 2AS `0-20^2=2a xx50` `F=-4ms^(-2)` Force of friction, F=ma=-4N |
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