A smooth wedge of mass M is pushed with an acceleration `a=gtantheta` and a block of mass m is projected down the slant with a velocity v relative to the wedge.
`The horizontal force applied on the wedge is:

Options

A. `6.37 m//s^(2)`
B. `12 m//s^(2)`
C. `10 m//s^(2)`
D. 5.77

Answer

C

Explanation

`F + N sin theta = MA` `mg - N cos theta = m a_(r ) sin theta`
`N sin theta = m (a_(r ) cos theta - A)` `W = Mg = 100 N implies M = 10 kg`
Solve to get: `a_(r ) = 20 m//s^(2)` and `A = 10 sqrt(3) m//s^(2)`
and acceleration of a 2kg block w.r.t ground is `sqrt(A^(2) + a_(r )^(2) + 2 Aa_(r ) cos (180^(@) - 30^(@))) = 10`