NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A conveyor belt is moving at a constant speed of 2 m s − 1 . A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 m s − 2 , is
Options
- A1 . 2   m
- B0 . 6   m
- CZero
- D0 . 4   m
Correct answer
D. 0 . 4   m
Step-by-step solution
The frictional force on the box f = μ m g ∴ Acceleration in the box a = μ g = 5   m   s − 2 v 2 = u 2 + 2 a s 4 = 0 + 2 × 5 × s s   =   0 . 4   m