An inclined plane of weight 100N and angle of inclination 30° is acted upon by a horizontal force of `100 sqrt(3 N)` as shown. A block of mass 2kg was kept on the inclined plane initially. Assume frictionless contacts. The acceleration of mass 2 kg w.r.t. ground is: (Take g = 10 m/`s^(2)`)

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`