Question

A 40kg slab rests on a frictionless tloor, A 10kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is0.40. The 10 kg block is acted upon by a horizontal force of 100n. If g = 9.8 m/s squir , the resulting accelerationof the slab will be also option
a. 1.47 m/s squir b. 1.69 m/s squir c. 9.8 m/s squir d. 0.98 m/s squir

Answer 1

maximum static friction force by slab on block of 10 kg
     = 0.60 * 10 kg * 9.8 m/s/s = 58.8 Newtons.

Since a force of 100 Newtons is applied, the block moves horizontally.  Since there is a kinetic friction between the two blocks. 

Horizontal net Force on the slab = kinetic friction between slab and block
       = 0.40 * 10 kg * 9.8 m/s/s = 39.2 Newtons

Acceleration of the slab = 39.2 N / 40 kg = 0.98 m/s/s

net force on the block = applied force - kinetic friction
acceleration of the block = (100 N - 39.2N) / 10 kg  = 6.18 m/s/s