A rough inclined plane BCE of height `(25/6)m` is kept on a rectangular wooden black ABCD of height 10 m , as shown in the figure. A small block is allowed to slide down from the top E of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is `1/8` and the angle of inclination of the inclined plane is `sin^-1 (0.6)`. IF the small block finally reaches the ground at a point F, then DF will (Acceleration due to gravity, `g=10 ms^-2`)

Options

A. `5/3 m`
B. `10 /3 m`
C. `13/3 m`
D. `20/3 m`

Explanation

According to question, a small block is slide down from top E of inclined plane as shown in figure,

Force equation of a block
`implies mg sin theta-f -ma`
`because` friction force applied one block `f= mu_k R`
or `f= mu_k R (mg cos theta)` ( From figure)
Where, `mu_k` = coefficient of kinetic friction
from Eq (i) we get
`implies mg sin theta- mu_k mg cos theta=ma`
`impliesa=10sin theta-1//8 times 10 cos theta.....(ii) (becausemu_k=1/8, given)`
`because` Given angle of the inclined plane, `theta= sin^-1 (0,6)`
or `sin theta=0.6`
`because cos theta=sqrt(1- sin ^2 theta)=sqrt(1-(0.6)^2)`
`therefore co stheta=0.8`
From Eq (ii)
`a=10 (0.6) -1/8 times 10 (0.8) or a=5 ms^-2`
When, block reached to point C, then from third equation of the motion,
`v^2=u^2+2as`
Where v= final velocity of the block at point C u= initial velocity of block at point E
or `v=sqrt(2as) ( because u=0)`
From `Delta EBC, EC= (BE)/ sin thetaimplies s=(25//6)/0.6=25/(6 times0.6)`
or`s=125/18 m`
Hence, `v=sqrt(2 times5 times125/18)=25/3ms^-1`
From second equation of motion of the block at point C,

In y- direction
`implies h= ut +1/2 gt^2`
`implies 10= u sin theta t+1/2 gt^2`
`implies 10=25/3 times 0.6t +1/2 times 10 times t^2`
`implies25/3times6/10t+1/2times10timest^2=10`
`implies5t+5t^2=10`
`impliest^2+t-2=0 or t=1 sec`
Now, again from second Eqs. of motion in x-direction
`implies DF=v cos thetat+0 or DF=25/3 times 0.8=20/3 m`