Question

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 1

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`