A small body of mass m is located on a horizontal plane at the point O. The body of mass m is located on a horizontal plane at the point O. The body acquires a horizontal velocity `v_0`. Find,
(a) the mean power developed by the friction force during the whole time of motion, if the friction coefficient `k=0.27`, `m=1.0kg`, and `v_0=1.5m//s`,
(b) the maximum instantaneous power developed by the friction force, if the friction coefficient varies as `k=alphax`, where `alpha` is a constant, and x is the distance from the point O.

Explanation

Let the body m acquire the horizontal velocity `v_0` along positive x-axis at the point O.
(a) Velocity of the body t seconds after the beginning of the motion,
`vecv=vecv_0+vecwt=(v_0-kg t)veci` (1)
Instantaneous power `P=vecF*vecv=(-kmgveci)*(v_0-kgt)veci=-kmg(v_0-kgt)`
From Eq. (1), the time of motion `tau=v_0//kg`
Hence sought average power during the time of motion
`lt P gt =(underset(0)overset(tau)int-kmg(v_0-kgt)dt)/(tau)=-(kmgv_0)/(2)=-2W` (On substitution)
From `F_x=mw_x`
`-kmg=mw_x=mw_x(dv_x)/(dx)`
or, `v_xdv_x=-kgdx=-agxdx`
To find `v(x)`, let us integrate the above equation
`underset(v_0)overset(v)intv_xdv_x=-alphagunderset(0)overset(x)intxdx` or, `v^2=v_0^2-alphagx^2` (1)
Now, `vecP=vecF*vecv=-malphaxgsqrt(v_0^2-alphagx^2)` (2)
For maximum power, `(d)/(dt)(sqrt(v_0^2x^2-lambdagx^4))=0` which yields `x=(v_0)/(sqrt(2alphag))`
Putting this value of x, in Eq. (2) we get,
`P_(max)=-1/2mv_0^2sqrt(alphag)`