A 40 kg slab rests on a frictionless floor . A 10 kg block rests on top of the slab (Fig . 7. 121). The static coefficient of friction between the block and the slab is `0.60` while the kinetic coefficient is `0.40` . The 10 kg block is acted upon by a horizontal force of 100 N . if g = `9.8 m//s^(2)` the resulting acceleration of the slab will be :

Options

A. `0. 98 m//s^(2)`
B. `1.47 m//s^(2)`
C. `1.52 m//s^(2)`
D. `6.1 m // s^(2)`

Answer

A

Explanation

Acceleration of slab
`a = (f_(K))/(M) = (mu_(k)N)/(M)`
= `(4)/(10) xx (10 xx 9.8)/(40) = 0.98 m // sec^(2)`