| ⇦ | 
| ⇨ |
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: - A system consists of three masses m₁, m₂ and m₃ connected by a string passing
- Source S₁ is producing, 10¹⁵ photons per second of wavelength 5000Å
- A ship A is moving westwards with a speed of 10 km/h and a ship B 100 km
- In Young’s double slit experiment, when wavelength used is 6000Å and the screen
- A bar magnet having a magnetic moment of 2 x 10⁴JT⁻¹ is free to rotate in a horizontal
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A system consists of three masses m₁, m₂ and m₃ connected by a string passing
- Source S₁ is producing, 10¹⁵ photons per second of wavelength 5000Å
- A ship A is moving westwards with a speed of 10 km/h and a ship B 100 km
- In Young’s double slit experiment, when wavelength used is 6000Å and the screen
- A bar magnet having a magnetic moment of 2 x 10⁴JT⁻¹ is free to rotate in a horizontal
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply