# 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

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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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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