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. `1.5m//s^(-2)`
B. `2.0m//s^(-2)`
C. `10m//s^(-2)`
D. `1.0m//s^(-2)`

Answer

D

Explanation

`f_(ms)=0.6xx10xx10N=60N`
Since the applied force is dreater than `f_(ms)` therefore the block will be in motion . So, we should consider `f_(k)`
`f_(k)=0.4xx10xx10N=40N`
This would cause acceleration of `40kg` block
Acceleration `=(4xx10)/(40kg)=1.0ms^(-2)` .