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A conveyor belt is moving at a constant speed of 2 m/s. 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 ms⁻², is
Options
(a) 1.2 m
(b) 0.6 m
(c) zero
(d) 0.4 m
Correct Answer:
0.4 m
Explanation:
Frictional force oon the box f = µmg
Acceleration in the box a = µg = 5 ms⁻²
v² = u² + 2as
⇒ 0 = 2² + 2 x (5)s
⇒ s = – 2/5 w.r.t. belt
⇒ distance = 0.4 m
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The time period of a mass suspended from a spring is T. If the spring is cut into four
- In cyclotron, for a given magnet, radius of the semicircle traced by positive ion
- An atom bomb works on the principle of
- A spring balance is attched to the ceiling of lift.A man hangs his bag on the string
- Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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