A 40 kg slab rests on frictionless floor as shown in fig. A 10 kg 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 10 kg block is acted upon by a horizontal force of 100 N. If `g=9.8m//s^(2)`, the resulting acceleration of the slab will be:

Options

A. `0.98m//s^(2)`
B. `1.47m//s^(2)`
C. `1.52m//s^(2)`
D. `6.1m//s^(2)`

Answer

A

Explanation

Force on the slab (m=40 kg) = reaction of frictional force by the upper block on the lower slab

`:.40a= mu_(k)xx10xxg or a=0.98 m//sec^(2)`