A 40kg slab rests on a frictionless floor as shown in the figure. A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is `0.60` while the kinetic friction is `0.40` . The 10kg block is acted upon by a horizontal force 100N. if `g=9.8m//s^(2)` , the resulting acceleration of the slab will be.

Options

A. `1 m//s^(2)`
B. `1.5 m//s^(2)`
C. `2 m//s^(2)`
D. `2.5 m//s^(2)`

Answer

A

Explanation

`f_(max) = mu_(s) N = 0.6 xx 100 = 60 N`
`100 N gt 60 N` , i.e. the upper block is in motion with respect to the lower block
`f_(k) = mu_(k) N = 0.4 xx 100 = 40 N`
`f_(k) = 40 a`
`40 = 40 a`
`a = 1 m//s^(2)`