A block slides down a frictionless plane inclined at an angle `theta`. For what value of angle `theta` the horizontal component of acceleration of the block is maximum? Find this maximum horizontal acceleration.

A block mass 'm' is placed on a frictionless inclined plane of inclination theta with horizontal. The inclined plane is accelerated horizontally so that the block does not slide down. In this situation vertical force exerted by the inclined plane on the block is:

As an inclined plane is made slowly horizontal by reducing the value of angle theta with horizontal. The component of weight parallel to the plane of a block resting on the inclined plane-

A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

A block of mass m is placed on a rough plane inclined at an angle theta with the horizontal. The coefficient of friction between the block and inclined plane is mu . If theta lt tan^(-1) (mu) , then net contact force exerted by the plane on the block is :

A block of mass m is placed on a rough plane inclined at an angle theta with the horizontal. The coefficient of friction between the block and inclined plane is mu . If theta lt tan^(-1) (mu) , then net contact force exerted by the plane on the block is :