A solid block of mass `2 kg` is resting inside a cube shown in figure The cube is moving with a velocity `hatv= 5 hati + 2hatj ms^(-1)` If the coefficient of friction between the surface of cube and block is `0.2` then the force of friction between the block and cube is

A soild block of mass 2 kg is resting inside a cube as shown in the figure .The cube is moving with a veocity v=5thati +2hatj m//s Here , t is is time in second the block is at rest with respect to the cube and cefficient of frision between the surface of ube and block is 0.6 then ,[take g=10m//s^(2) ]

A hollow cubical box P is moving on a smooth horizontal surface in the x-y plane with constant acceleration of vec(a) = 3 hat (i) + 4 hat (j) m//s^(2) .A block Q of mass 2 kg is at rest inside the cubical box as shown in figure. If the coefficient of friction between the surface of the cube P and the block Q is 0.6. Then the force of friction between P and Q is :

A hollow cubical box P is moving on a smooth horizontal surface in the x-y plane with constant acceleration of vec(a) = 3 hat (i) + 4 hat (j) m//s^(2) .A block Q of mass 2 kg is at rest inside the cubical box as shown in figure. If the coefficient of friction between the surface of the cube P and the block Q is 0.6. Then the force of friction between P and Q is :

In the arrangement shown in figure, coefficient of friction between the two blocks is mu=1//2 . The force of friction acting between

A solid sphere of mass 2 kg is resting inside a cube as shown in the figure . The cube is moving with a velocity v=5ti + 2tj . Here t is in second . All surface are smooth . The sphere is at rest with respect to the cube . What is the total force exerted by the sphere on the cube .

A solid sphere of mass 2 kg is resting inside a cube as shown in the figure . The cube is moving with a velocity . Here t is in second . All surface are smooth . The sphere is at rest with respect to the cube . What is the total force exerted by the sphere on the cube .

A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface and the block is 1/4 .Then the friction force acting on the block is

A block of mass 4kg is placed in contact with the front vertical surface of a lorry.The coefficient of friction between the vertical surface and block is 0.8 .The lorry is moving with an acceleration of 15m//s^(2) .The force of friction between lorry and block is (g=10ms^(-2))

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A force F of2.5 N is applied on the block as shown. Calculate the force of friction between the block and the floor