A block is moving on an inclined plane making an angle `45^@` with the horizontal and the coefficient of friction is `mu`. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define `N=10mu`, then N is

Options

A. `3`
B. `4`
C. `5`
D. `6`

Answer

c

Explanation

Force requied to just push the block up the inclined plane of inclination `theta` is
`F_(up) = mg sin theta + mu mg cos theta`
Force required to just prevent the block from sliding down s
`F_(down) = mg sin theta - mu mg cos theta`
According to question `F_(up) = 3 F_(down)`
`:. mg sin theta + mu mg cos theta = 3 (mg sin theta - mu mg cos theta)`
or `sin theta + mu cos theta = 3 (sin theta - mu cos theta)`
As `theta =45^(@)` therefore
`(1)/sqrt2 + mu xx(1)/sqrt2=3[(1)/sqrt2-muxx(1)/sqrt2]`
On solving `mu =(1)/(2)`
Given `N = 10 mu = 10 xx (1)/(2) = 5` .