A block of mass `m` is placed on a rough inclined plane. When the inclination of the plane is `theta` , the block just beging to slide down the plane under its own weight. The minimum force applied parallel to the plane, to move the block up the plane, is.

A block of mass 10 kg is placed on an inclined plane when the angle of inclination is 30^(@) . The block just begins to slide down the plane. The force of static friction is

A block of mas M is resting on an inclined plane as shown in the figure. The inclination of the plane to the horizontal is gradually increased. It is found that when the angle of inclination is theta the block just begins to slide down the plane. What is the minimum force F applied parallel to the plane that would just make the block move up the plane ?

A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased till the block just begins to slide down. The inclination of the plane is than 3 in 5. The coefficient of friction between the block and the plane is (Take, g = 10m//s^(2) )

A block is placed on an inclined plane. The angle of inclination of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is

A block of mass m is placed on a smooth inclined plane of inclination theta with the horizontal. The force exerted by the plane on the block has a magnitude

A block of mass m is at rest on a rough inclined plane of angle of inclination theta . If coefficient of friction between the block and the inclined plane is mu , then the minimum value of force along the plane required to move the block on the plane is

A block of mass 5 kg is at rest on a rough inclined surface. If angle of inclination of plane is 60^(@) , then force applied by it on block is

A block is about to slide down an inclined plane when its inclination to the horizontal is theta . If now a 5 kg weight is attached on the block :