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) )

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))

Two blocks of mass 4kg and 2kg are connected by a heavy string and placed on rough horizontal plane. The 2 kg block is pulled with a constant force F. the coefficent of friction between the blocks and the ground is 0.5. what is the value of F so that tension in the string is constant throughout the length? (g=10m//s^(2))

In the figure, what should be mass m so that block A slides up with a constant velocity ?

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

A stress of 10^(6) N//m^(2) is required for breaking a material. If the density of the material is 3 xx 10^(3) Kg//m^(3) , then what should be the minimum length of the wire made of the same material so that it breaks by its own weight (g = 10m//s^(2))

Two smooth blocks of same mass are connected by an inextensible and massless string which is passing over a smooth pulley are kept in a lift is going down with acceleration 'a' as shown in the fig What should be the value of a (in m//s^(2)) so that acceleration of block A w.r.t. ground will be minimum? (g=10m//s^(2))