Find M (in kg ) for which `m_(2)` will start sliding over `m_(1)` .Table on which `m_(1)` and `m_(2)` are placed is smooth,`mu` is the friction coefficient between `m_(1)` and `m_(2)` . (Given :`m_(1)`=2kg, `m_(2)`=1kg, `mu`=0.5, `g=10 m/s^(2)`)

m_(A) = 1 kg, m_(B) = 2 kg, m_(C) = 10 kg

In the figure shown m_(1)=5 kg,m_(2) = 10 kg & friction coefficient between m_(1) & m_(2) is mu=0.1 and ground is frictionless then :

In the system of connecting blocks, m_(A) =1kg, m_(B)=2kg and m_(C)=2kg .The inclined plane is smooth while the friction coefficient between block B and horizontal ground is mu .Mark the correct options.(g=10m s^(-2) )

Two masses m_(1) and m_(2) are attached to a string which passes over a frictionless smooth pulley. When m_(1)=10kg,m_(2)=6kg , the acceleration of masses is

There are two blocks of masses m_(1) and m_(2) , m_(1) is placed on m_(2) on a table which is rotating with an angular velocity omega about the vertical axis . The coefficent of friction between the block is mu_(1) and between m_(2) and table is mu_(2) (mu_(1) lt mu_(2)) if block are placed at distance R from the axis of ratation , for relative sliding between the surface in contact , find the a. friction force at the contacting surface b. maximum angular speed omega

A bullet of mass 0.25 kg is fired with velocity 302m/s into a block of wood of mass 37.5kg.It gets embedded into it. The block m_(1) is resting on a long block and the horizontal surface on which it is placed. The coefficient of friction between m_(1) and m_(2) is 0.5. Find the displacement of on and the common velocity of m_(1) and m_(2) . Mass m^(2) = 12.5kg.

A board of mass m = 1 kg lies on a table and a weight of M = 2 kg on the board. What minimum force F must be applied on the board in order to pull it out from under the road ? The coefficient of friction between board and table is mu_(2) = 0.5 : (Take g = 10 m//s^(2) )