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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
- An object is placed at a distance of 10 cm from a co-axial combination of two lenses
- The angle between the two vectors A = 3i+4j+5k and B = 3i+4j-5k will be
- In insulators (CB is Conduction Band and VB is Valence Band)
- In a circuit, L,C and R are connected in series with an alternating voltage source
- The velocity of a particle at an instant is 10 ms⁻¹ and after 5 s the velocity
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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