A 40 kg slab rests on a frictionless floor . A 10 kg block rests on the top of the slab . [Fig. 7.81 (a)] . The coefficients of static and dynamic friction are `0.60` and `0.40` respectively . The 10 kg block is pulled by a horizontal force of 100 N . what are the resulting accelerations of (a) the block (b) the slab ? ( g = `10 m// s^(2)`)

Answer

`a_(B) = 6 m//s^(2)` and `a_(S) = 1 m//s^(2)`

Explanation

If they move with common acceleration then ,
`a = (100)/(40 + 10) = 2 m//s^(2)`

Pseudo force on block = `10 xx 2 = 20 N`
Net force = `100 -20 = 80 N`
`f_("s max") = mu_(s) xx 10 xx 10 = 60 N`
`therefore 80 N gt 60N `
there will be relative motion .
Now , FBD of block and slab will be

`therefore 100 - f_(k) = 10 a_(1) " " f_(k) = 40 a_(2)`
`100 - 0.4 xx 10 xx 10 = 10 a_(1) " " 0.4 xx 10 xx 10 = 40 a_(2)`
`therefore " " a_(1) = 6 m//s^(2) therefore a_(2) = 1 m//s^(2)`