A man wants to slide down a block of mass `m` which is kept on a fixed inclined plane of inclination `30^(@)` as shown in the figure .Initially the block is not sliding .
To just start sliding the man pushed the block down the inclined with a force `F`. Now , the block starts accelerating. To move it downwards with conatant speed the man starts pulling the block with same force. Surface are such that ratio of maximum static friction to kinetic friction is `2` Now, answer the following questions.

What is the value of `F`?

Options

A. `(mg)/(4)`
B. `(mg)/(6)`
C. `(mg sqrt3)/(4)`
D. `(mg)/(2 sqrt3)`

Answer

B

Explanation

Let `mu_(k) = mu` then `mu_(s) = 2 mu`
According to first condition, `F + mg sin theta = mu_(s) mg cos theta = 2 mu mg cos theta` …(i)
According to second condition,
`mg sin theta = F + mu_(k) mg cos theta`
`= F + mu_ mg cos theta` ...(ii)
Putting `theta = 30^(@)`, we get
`F + mg// 2 = 2 mu mg ((sqrt3)/(2))`
or `sqrt3 mu mg = F + 0.5 mg` ...(iii)
`(mg)/(2) = F + mu mg ((sqrt3)/(2))`
or `0.5 sqrt3 mu mg = 0.5 mg - F` ...(iv)
Driving Eq. (iii) and (iv) , we get
`F = (mg)/(6)`