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  • 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

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