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. `6.1 ms^(-2)`
B. `1.52 ms^(-2)`
C. `1.47 ms^(-2)`
D. `0.98 ms^(-2)`

Answer

D

Explanation

(d)Force of limiting friction for block `=mu_(s)mg`
`=0.60xx10xx9.8=58.8 N`
If the applied force is greater than 58.8 N, then the block will move over the slab.
Kinetic friction acition on the block towards right
`=mu_(K)mg=0.40xx10xx9.8=39.2 N`
This is also equal to the force of friction acting on slab towards left.
This is only force acting on slab.
So, acceleration of the slab, `a=(F)/(m)=(39.2)/(40)=0.98 ms^(-2)`