A 20 kg metal block is placed on a horizontal surface. The block just begins to slide when horizontal force of 100N is applied to it . Calculate the coefficient of static friction . If coefficient of kinetic friction is 0.4 then find minimum force to maintain its uniform motion.

A block of 50 kg rests on a table. A horizontal force of 294 N is required to just move the block, the coefficient of static friction between the surfaces in contant (mu_(s)) is

A block of mass 10kg is placed on a rough horizontal surface having coefficient of friction mu=0.5 . If a horizontal force of 100N is acting on it, then acceleration of the will be.

A block of mass 4 kg resting on horizontal surface can be kept in uniform motion on the horizontal surface by the application of the force 16 N, the coeffcient of kinetic friction between the two surfaces is

If a force of 0.98 N is applied to a mass of 0.25 kg (g = 9.8m//s^2) the coefficient of friction is

The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.

The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.

A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude A. What will be the frequency of oscillation, the block will just start to slip? Coefficient of friction =mu .

A block A with mass 100kg is resting on another block B of mass 200kg. As shown in figure a horizontal rope tied to a wall hold it. The coefficient of friction between A and B is 0.2 while coefficient of friction between B and the ground is 0.3 . the minimum required force F to start moving B will be.

A metal plate of area 500 cm^92) is kept on a horizontal surface with a layer of oil of thickness 0.5 mm between them. The horizontal force required to drag the plate with a velocity of 2 cm/s is (coefficient of viscosity =0.9 kg/ms)

The coefficient of static friction between a block of mass 0.25 kg is 0.4. Calculate the force applied to it. (g = 9.8m//s^2)