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).

A small body of mass m moving with velocity v_(0) on rough horizontal

1.	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).
Answer» Solution :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`