| ⇦ |
| ⇨ |
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: - Curie temperature above which
- An insulated container of gas has two champers seperated by an insulating partition.
- Efficiency of a Carnot engine is 50% when temperature of sink is 500K. In order
- On a frictionless surface, a block of mass M moving at speed v collides elastically
- An inductor coil is connected to a 12 V battery and drawing a current 24 A. This coil
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Curie temperature above which
- An insulated container of gas has two champers seperated by an insulating partition.
- Efficiency of a Carnot engine is 50% when temperature of sink is 500K. In order
- On a frictionless surface, a block of mass M moving at speed v collides elastically
- An inductor coil is connected to a 12 V battery and drawing a current 24 A. This coil
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