A small body of mass `m` is given velocity `v_0` at point `O` on rough horizontal surface. (a) Find the mean power developed by the friction force during the whole time of motion if the friction coefficient is constant and equal to `mu_0` (b) Find the maximum instantaneous power developed by the friction force, if the friction coefficient varies as `mu=mu_0x`, where `mu_0` is a constant and `x` is the distance from the point `O`.

Explanation

(a) Loss in `K.E.=`work done against friction
`(1)/(2)mv_0^2=W_f`
Work done by friction `W_f^(')=-W_f=-(1)/(2)mv_0^2`
Retardation of body `a=(f)/(m)=(mu_0mg)/(m)=mu_0g`
The time after which the body stops
`0=v_0-at=v_0-mu_0gtimpliest=(v_0)/(mu_0g)`
(b) Retardation of body
`a=(f)/(m)=(mumg)/(m)=mug=mu_0gx`
here retardation is variable Velocity of body as a function of `x`
`v(dv)/(dx)=-mu_0gx`
`int_(v0)^(v)dv=-mu_0gint_(0)^(x)xdx`
`(v^2)/(2)-(v_0^2)/(2)=-(mu_0gx^2)/(2)impliesv=sqrt(v_0^2-mu_0gx^2)`
Instantaneous power delevered by friction `P=vecF.vecF=-fv`
`P=-fv=-mu_0mgxsqrt(v_0^2-mu_0gx^2)`
`P^2=(mu_0mg)^2(v_0^2x^2-mu_0gx^4)`
For `P` to be maximum,
`(dP)/(dx)` or `(d(P^2))/(dx)=0`
`v_0^2xx2xx-mu_0gxx4x^3=0impliesx=(v_0)/(sqrt(2mu_0g))`
`P_(max)=-mu_0mg(v_0)/(sqrt(2mu_0g))sqrt(v_0^2-mu_0g(v_0^2)/(2mu_0g))`
`=-(1)/(2)mv_0^2sqrt(mu_0g)`