The `10 kg` block is moving to the left with a speed of `1.2 m//s` at time `t = 0` A force `F` is applied kinetic friction is `mu_(k) = 0.2`. Determine the time `t` at which the block comes to a slip. `(g = 10 m//s^(2))`

A block 'A' of mass 1kg is kept on a block 'B' of mass 2kg .Block A is imparted a velocity of 16m/s at t=0.If the coefficient of friction between A and B is 0.2, find the time (in sec) after which momentum of A and B relative to ground will be equal.(Take 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 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))

In the given figure, the blocks of mass 2kg and 3kg are placed one over the other as shown. The surface are rough with coefficient of friction mu_(1)=0.2, mu_(2)=0.06 . A force F=0.5t (where 't' in sec) is applied on upper block in the direction shown. Based on above data answers the following questions. (g = 10 m//sec^(2)) The frictional force acting between 3 kg block and ground w.r.t. time will vary as :

In the given figure, the blocks of mass 2kg and 3kg are placed one over the other as shown. The surface are rough with coefficient of friction mu_(1)=0.2, mu_(2)=0.06 . A force F=0.5t (where 't' in sec) is applied on upper block in the direction shown. Based on above data answers the following questions. (g = 10 m//sec^(2)) The frictional force acting between the blocks at t=10 sec is

In the given figure, the blocks of mass 2kg and 3kg are placed one over the other as shown. The surface are rough with coefficient of friction mu_(1)=0.2, mu_(2)=0.06 . A force F=0.5t (where 't' in sec) is applied on upper block in the direction shown. Based on above data answers the following questions. (g = 10 m//sec^(2)) The friction force between the blocks and time graph is :

Two blocks M_1 . and M_2 , are tied with string and kept at rough surface as shown in figure. If F = (2t)N is applied on the block of mass M_1 , then the friction force on the blocks at t = 2 s. is

In the given figure, the blocks of mass 2kg and 3kg are placed one over the other as shown. The surface are rough with coefficient of friction mu_(1)=0.2, mu_(2)=0.06 . A force F=0.5t (where 't' in sec) is applied on upper block in the direction shown. Based on above data answers the following questions. (g = 10 m//sec^(2)) The realtive slipping between the blocks occurs at t=

In the given figure, the blocks of mass 2kg and 3kg are placed one over the other as shown. The surface are rough with coefficient of friction mu_(1)=0.2, mu_(2)=0.06 . A force F=0.5t (where 't' in sec) is applied on upper block in the direction shown. Based on above data answers the following questions. (g = 10 m//sec^(2)) The frictional force acting between the two blocks at t=9 sec