Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A block of mass ‘ m ‘ is attached to a spring in natural length of spring constant ‘ k ‘ . The other end A of the spring is moved with a constant velocity v away from the bl

A block of mass ‘ m ‘ is attached to a spring in natural length of spring constant ‘ k ‘ . The other end A of the spring is moved with a constant velocity v away from the block . Find the maximum extension in the spring.

                                                          

Option: 1

\frac{1}{4}\sqrt{\frac{m\: v^{2}}{k}}        


Option: 2

\sqrt{\frac{m\: v^{2}}{k}}


Option: 3

\frac{1}{2}\sqrt{\frac{m\: v^{2}}{k}}


Option: 4

2\sqrt{\frac{m\: v^{2}}{k}}


Answers (1)

best_answer

As we have learnt in

 

Potential Energy stored in the spring -

U= \frac{1}{2}\: kx^{2}

- wherein

K=spring \: constant

x= elongation or compression of spring from natural position

 

  During maximum extension, the block will come to rest  

  Now, by energy conservation,

\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}_{max} \; \; \; \; \therefore X_{max}=\sqrt{\frac{m\: v^{2}}{k}}

 

Posted by

Ritika Kankaria

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

NTA Overhaul: 1,000 secure exam centres, biometrics to prevent fraud, question paper changes, suggests panel
anam-khan

NEET syllabus 2025 out; NMC uploads subject-wise topics
anam-khan

What changes in NEET UG? Experts’ panel suggests multi-stage exam, security overhaul, simpler process to NTA

Latest Question