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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
Related Questions: - Which phenomenon best supports the theory that matter has a wave nature?
- If μᵥ=1.5230 and μʀ=15.145, then dispersive power of crown glass is
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Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which phenomenon best supports the theory that matter has a wave nature?
- If μᵥ=1.5230 and μʀ=15.145, then dispersive power of crown glass is
- Two identical long conducting wires AOB and COD are placed at right angle
- The energy of groundstate (n=1) of hydrogen level is -13.6 eV. The ionistation
- The near point and far point of a person are 40 cm and 250 cm, respectively.
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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