An inclined plane is moving with Constant velocity `v = 4 m//s` on a horizoontal surface as shown in figure. If a block of mass `2 kg` is kept at top of the incline and their is no friction between the block and the incline, then the distance travelled by the incline till the block reaches bottom of the inclined is `(g = 10 m//s^(2))`

Options

A. `4 m`
B. `5 m`
C. `1.6 sqrt5 m`
D. `2 sqrt5 m`

Answer

D

Explanation

`a = g sin53^(@) = (4g)/(5)` time taken by block to reach the bottom
`:. t = sqrt((25)/(a)) = sqrt((25)/(2 xx 10)) = sqrt((5)/(4))`
`:.` Distance travelled by the inclined `= vt = 4sqrt((5)/(4)) = 2 sqrt5 m`