A conveyor belt is inclined at an angle of 30^ to the horizontal and moves upwards at a constant speed of 6 m s ⁻¹ . A block is gently placed on the belt and slips for some time before catching up to the belt's speed. If the coefficient of kinetic friction between the block and the belt is 3 , the distance travelled by the conveyor belt during the time the block slips on it is: [Take g = 10 m s ⁻² ]

Step-by-step solution

When the block is gently placed on the moving belt, its initial velocity is zero. The kinetic friction acts upwards along the incline to accelerate the block, while the component of gravity acts downwards along the incline. The net acceleration of the block up the incline is given by: a = g - g Substituting the given values: a = 3 10 30^ - 10 30^ a = 10 3 ( 3 2 ) - 10 ( 1 2 ) = 15 - 5 = 10 m s ⁻² The time taken for the block to reach the belt's speed ( v = 6 m s ⁻¹ ) is: t = v a = 6 10 = 0.6 s The distance travelle