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 `0.04` . 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. `1m//s^(2)`
B. `1.5m//s^(2)`
C. `2m//s^(2)`
D. `6m//s^(2)`

Answer

A

Explanation

Limiting friction between block and slab `=mu_(s)m_(A)g`
`=0.6xx10xx10=60N`
But applied force on block `A` is `100N` . So the block will slip over a slab.
Now kinetic friction works between block and slab`Fk=mu_(k)m_(A)g=0.4xx10xx10=40N`
This kinetic friction helps to move the slab
:. Acceleration of slab `=(40)/(m_(B))=(40)/(40)=1m//s^(2)` .