What should be minimum value of F so that 2 kg slides on ground but 1kg does not slide on it? `[g = 10m/sec^(2)]`

What should be the maximum value of M so that the 4 kg block does not slip over the 5 kg block : (Take g=10 m//s^(2) )

A block of mass m=2kg is resting on a rough inclined plane of inclination of 30^(@) as shown in figure. The coefficient of friction between the block and the plane is mu=0.5 . What minimum force F should be applied perpendicular to the plane on the block, so that blocks does not slip on the plane? (g=10m//s^(2))

Initially, both the blocks are at rest on horizontal surface as shown in the figure. Find the minimum value of force F in (N) so that sliding starts between the blocks (g=10ms^(-2))

Initially, both the blocks are at rest on horizontal surface as shown in the figure. Find the minimum value of force F in (N) so that sliding starts between the blocks (g=10ms^(-2))

A body moves along a circular path of radius 1m & mu = 0.4 . The maximum angular velocity in rad/sec of the body so that it does not slide is (g = 10 ms^(-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)

Initially spring is in natural length and block of mass m is released from rest what should be the minimum value of M (in kg) to just move the 2kg block. If this minimum value is (x)/(8) kg. Find x

Two blocks a and B of mass m_(A)=1kg and m_(B)=3kg are kept on the table as shown in figure the coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 the maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is [Take g=10m//s^(2) ]