Find the maximum possible force which can be applied to the 8 kg block shown in figure-2.94 to move both the blocks together if bottom surface is (a) frictionless, (b) having friction coefficient 0.3. Take `g = 10m//s^(2)`

The maximum value of F which can be applied on the system shown in figure so that both blocks move together with same acceleration is

Find the minimum force F_0 that should be applied to the block of mass m = 0.5 kg so that the block stays in equilibrium with the rough vertical wall having friction coefficient mu = 0.2 (Take g = 10 m/s^2 )

A force of 100N is applied on a block of mass 3 kg as shown in the figure-2.164. The coefficient of friction between the surface and the block is 0.25.The frictional force acting on the block is: (Take g=10m/s^(2) )

A 2 kg block is present against a rough wall by a force F = 20 N as shown in figure .Find acceleration of the block and force of friction acting on it . (Take g = 10 m//s^(2))

A force of 200N is applied as shown in the figure on block of 3 kg. The coefficient of friction between the block and the wall is 0.3. Find the friction acting on the block. (in N) {Take g = 10 m//s^(2)

A block of mass m is placed on another block of mass M lying on a smooth horizontal surface as shown in the figure. The co-efficient of friction between the blocks is mu . Find the maximum horizontal force F (in Newton) that can be applied to the block M so that the blocks together with same acceleration? (M=3kg, m=2kg, mu=0.1, g=10m//s^(2))

A block of mass 10 kg is placed on a horizontal surface as shown in the figure and a force F = 100 N is applied on the block as shown, in the figure. The block is at rest with respect to ground. If the contact force between block and ground is 25n (in Newton) then value of n is : (Take g = 10 m//s^(2))

A block of mass 6 kg is kept on rough surface as shown in figure. Find acceleration and friction force acting on the block. (Take g=10m//s^(2) )

The block is always at rest, the maximum force which can be applied for this is 24x. Find the value of x. Take g = 10 m//s^(2) {coefficient of friction = 0.3}