Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • 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 <img alt="\mu=0.5" src="https://learn.careers360.com/latex-image/?

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 \mu=0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 ms-2, is

Option: 1

Zero


Option: 2

0.4 m


Option: 3

1.2 m


Option: 4

0.6 m


Answers (1)

best_answer

Force, F =\mu g$ \\ Retardation of the block on the belt \\ $a=\frac{F}{m}=\frac{\mu m g}{m}=\mu g$ \\ From, $v^{2}=u^{2}+2$ as $0=(2)^{2}-2(\mu g) s$ \\ $s=\frac{4}{2 \times 0.5 \times 10}=0.4 m$

 

Posted by

Gaurav

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET MDS Counselling 2025 round 2 results out; report by July 27
anam-khan

NEET Counselling 2025: MBBS fees skyrocket, cross Rs 1 crore at 32 deemed-university medical colleges
anam-khan

UP NEET UG counselling 2025 registration begins; 78 candidates debarred from MBBS, BDS admissions

Latest Question