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A stone of 1 kg is thrown with a velocity of `20ms^(-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» Here, Initial velocity, u=`20ms^(-1)` final velocity,v=0 (The stone stops) Acceleration,a = ? (To be calculated) And, Distance travelled, s=50 m Now, `v^(2)=u^(2)+2as` `(0)^(2)=(20)^(2)+2xxaxx50` 0-400+100 a 100 a=-400 `a=-400/100` `a=-4ms^(-2)` Now, Force,F=`mxxa` `F=1xx(-4)N` F=-4N Thus, the force of friction between the stone and the ice is 4 newtons. The negative sign shows that this force opposes the motion of stone. |
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