Friction coefficients at all surfaces in the given setup is `mu_(s)=0.5` `mu_(k)=0.20` .Find the minimum acceleration of the block of mass 2kg so that smaller block of mass 0.5kg remains at rest wrt 2kg (g=10m/s^(2))

A disc of radius R=2m starts rotating wth constant angular acceleration alpha=(2rad)//s^(2) . A block of mass m=2kg is kept at a distance (R)/(2) from the centre of disc. The coeffiecient of friction between disc and block is mu_(s)=0.4,mu_(k)=0.3 . Acceleration of block when it just slips w.r.t. ground and w.r.t. disc are respectively - ( g=10ms^(-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))

If coeffiecient of friction between the block of mass 2kg and table is 0.4 then out acceleration of the system and tension in the string. (g=10m//s^(2))

A system of two blocks is shown in figure. Friction coefficient between 5 kg and 10 kg block is mu=0.6 and between 10 kg and ground is mu=0.4 What will be the maximum value of force F applied at the lower block so that 5 kg block does not slip w.r.t. 10 kg . (g=10m//sec^(2)). The force applied at the upper block is having fixed magnitude of 80 N (both forces start to act simultaneously)

A block of mass 8 kg is sliding on a surface inclined at an angle of 45^(@) with the horizontal. Calculate the acceleration of the block. The coefficient of friction between the block and surface is 0.6. (Take g=10m//s^(2) )

A block of mass 8 kg is sliding on a surface inclined at an angle of 45^(@) with the horizontal . Calculate the acceleration of the block . The coefficient of friction between the block and surface is 0*6 ( Take g = 10 m//s^(2) )

Figure shown two blocks in contact sliding down an inclined surfase of inclination 30^(@) . The friction coefficient between the block of mass 4.0kg and the incline is mu _(2) =0.30 . Find the acceleration of 2.0kg block. (g = 10m//s^(2))

Figure shows two blocks in contact placed on an incline of angle theta = 30^(@) . The coefficient of friction between the block of mass 4 kg and the incline is m1, and that between 2 kg block and incline is mu_(2) . Find the acceleration of the blocks and the contact force between them if – (a) mu_(1) = 0.5 , mu_(2) = 0.8 (b) mu_(1) = 0.8 , mu_(2) = 0.5 (c) mu_(1) = 0.6 , mu_(2) = 0.1 " " [ Take g = 10m//s^(2) ]