A block mass `m` moves on a horizontal circle against the wall of a cylinderical room of radius `R`. The floor of the room, on which the block moves, is smooth but the friction coefficient between the wall and the block is `mu`. The block is given an initial speed `V_(0)`. the power developed by the resultant force acting on the block as a function of distance travelled `s` is

Options

A. `(mum_(0)^(3))/Re^((-3s)/(mu))`
B. `(mumV_(0)^(3))/Re^((-3mus)/R)`
C. `(mumV_(0)^(3))/R`
D. `(mumV_(0)^(3))/Re^((-3s)/R)`

Answer

B

Explanation

`f=-ma`
`rArrmuN=-ma`
`=rarr mu(mV^(2))/R=-mxxV(dV)/(dS)`
`rArr(muV)/R=(-dV)/(dS)`
`rArr(dV)/(V)=(-mu)/R dS`
`rArrint_(V_(0))^(V)(dV)/V=(-mu)/Rint_(0)^(S)dS`
`rArrln|V/(V_(0))|=(-mu)/R(S)`
`rArrV=V_(0)e^((-muS)/R)....(1)`
Now power consumed by friction,
`P=-f.V`
`=-muNV`
`=-mu(mV^(2))/RV`
`=-mu(mV^(3))/R......(2)`
substituting the value of `V` from eq (1), we get
`P=(-mum)/R V_(0)^(3)e^((-3muS)/R)`