A slab of `40 kg` lying on a frictionales floor and a block of 10 kg rests on the slab as shown in the figure. The coefficients of static and kinetic friction between the slab and the block are `0.6` and `0.4` respectively. If a force of `100 N` is applied on the block, then the acceleration the slab will be `(g = 10 m//s^(2))`

Block A of mass 5 kg is resting on a frictionless floor. Another block B of mass 7kg is resting on it as shown in the figure. The coefficient of friction between the blocks is 0.5 while kinetic friction is 0.4. If a force of 100 N is applied to block B, the acceleration of the block A will be ( g =10 ms^(-2)) :

A block of mass 5 kg is placed on a rough horizontal plane . The coefficients of stastic and kinetic friction between the block and surface are 0*5 and 0*3 respectively. A force of 100 N is applied at an angle 30^(@) with the horizontal as shown in the figure . The acceleration of the block will be [ g = 10 m//s^(2) ]

A block B of mass 5kg is placed on a slab A of mass 20kg which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is 0.4 and that of kinetic friction is 0.2. If a force F = 25N acts on B, the acceleration of the slab will be:

A block B of mass 5kg is placed on a slab A of mass 20kg which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is 0.4 and that of kinetic friction is 0.2. If a force F = 25N acts on B, the acceleration of the slab will be:

Consider the arrangement shown in the figure. The coefficient of friction between the two blocks is 0.5 and between the 4 kg block and the horizontal surface is 0.4. If a horizontal force of 12 N is applied on the 4 kg block, the acceleration of the 2 kg block would be

A 40kg slab rests on a frictionless floor. A 10kg block rests on top of the slab. the static coefficient of friction between the block and the slab is 0.60 and The coefficient kinetic friction between the block and the slab is 0.40. A horizontal force of 100N is applied on the 10kg block. Find the resulting acceleration of the slab. (g=10ms^(-1))