A force of 100N is applied on a block of mass 3kg as shown in the figure. The coefficient of friction between the surface and the block is `mu=(1)/(sqrt(3)` . The friction force acting on the bock is.

A force of 100N is applied on a block of mass 3 kg as shown in the figure-2.164. The coefficient of friction between the surface and the block is 0.25.The frictional force acting on the block is: (Take g=10m/s^(2) )

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 force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is 1//4 . The friction force acting on the block is :

A force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is 1//4 . The friction force acting on the block is :

A force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is 1//4 . The friction force acting on the block is :

If coefficient of friction between both the block is mu- (1)/(2) . Find friction force acting between both block.

A block of mass 6 kg is being pulled by force 24 N as shown . If coefficient of friction between the block and the surface is 0.6 , frictional force acting on the block is :

A force of 200N is applied as shown in the figure on block of 3 kg. The coefficient of friction between the block and the wall is 0.3. Find the friction acting on the block. (in N) {Take g = 10 m//s^(2)

A force of 200N is applied as shown in the figure on block of 3 kg. The coefficient of friction between the block and the wall is 0.3. Find the friction acting on the block. (in N) {Take g = 10 m//s^(2)

In the arrangement shown in figure coefficient of friction between 5kg block and incline plane is mu=0.5 . Friction force acting on the 5kg block is