InterviewSolution
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InterviewSolution
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A small body of mass m moving with velocity `v_(0)` on rough horizontal surface, finally stops due to friction. Find, the mean power developed by the friction force during the motion of the body, if the frictional coefficient `mu`=0.27,m=1 kg and `v_(0)=1.5 ms^(-1)`. |
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Answer» The retardation due to friction is, `a=("force of friction")/("mass")=(mu mg)/(m)=-mug` Further,`" " v_(0)`=at Therefore, `" "t=(v_(0))/(a)=(v_(0))/(mug)`………..(i) From work energy theorem, work done by force of friction =change in kinetic energy or `" " W=(1)/(2)mv_(0)^(2)`..............(ii) Mean power`=(W)/(t)` From Eqs. (i) and (ii), we get `P_(mean)=(1)/(2) mu mgv_(0)` Substituting the values, we have `P_(mean)=(1)/(2)xx0.27xx1.0xx9.8xx1.5~~2 W` |
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