A block of mass `m` is pulled slowly by a minimum constant force `(F)` on a horizontal surface thorugh a distance x. The coefficient of kinetic friciton is `mu`. Find the work done by the force `(F)`.

Options

A. `(mu mgx)/(1 + mu^(2))`
B. `(mu mgx)/(1 - mu^(2))`
C. `(mu mgx)/(sqrt(1 + mu^(2)))`
D. None of these

Answer

A

Explanation

The work done by the force `F` is
`W = F_(min) xx cos theta` ...(i)

`cos theta = (1)/(sqrt(1 + e^(2))) sin theta = (mu)/(sqrt(1 + e^(2)))`
`F` is minimum when `tan theta = mu` (ii)
`F_(min) = mg sin theta` (iii)
Using equation (i) and (iii).
`W = (mg sin theta) xx cos theta`
`= mg = (mu)/(sqrt(1 + mu^(2))) xx (1)/(sqrt(1 + mu^(2))) = (mu mgx)/((1 + mu^(2))`.