Sand is droped vetically downward at the constant rate `mu kg//s` on a conveyor belt which is moving horizontally with velocity `'v'`. Assume that sant particles comes to rest with respect to belt immediately after landing.
The rate which work is done by friction force on sand is

Sand is droped vetically downward at the constant rate mu kg//s on a conveyor belt which is moving horizontally with velocity 'v' . Assume that sant particles comes to rest with respect to belt immediately after landing. The rae at which work is done by the external force applied on the belt.

Sand is droped vetically downward at the constant rate mu kg//s on a conveyor belt which is moving horizontally with velocity 'v' . Assume that sant particles comes to rest with respect to belt immediately after landing. The magnitude of force must be applies on the belt moves with same constant velocity v .

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The extra force required is

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The extra force required to keep the belt moving is

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The extra force required to keep the belt moving is

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The extra force required to keep the belt moving is

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The time rate of change of kinetic energy of sand particles is

Sand particles drop vertically at the rate of 2kgs^-1 on a conveyor belt moving horizontally with a velocity of 0.2ms^-1 . The time rate of change of kinetic energy of sand particles is

Sand drops fall vertically at the rate of 2 kg/sec on to a conveyor belt moving horizontal with a velocity 0.2 m/s. Then the extra force required to keep the belt moving is