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  • I have a doubt, kindly clarify. - Work Energy and Power - JEE Main-3

 A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity v is :

  • Option 1) (1/2)mv2   
  • Option 2) mv2 
  • Option 3) (1/3)mv2   
  • Option 4) (1/6)mv2   
 

Answers (2)

best_answer

As we have learned

Kinetic energy -

k= \frac{1}{2}mv^{2}

- wherein

m\rightarrow mass

v\rightarrow velocity 

Kinetic Energy is always positive.

 

 

 fixed end is at rest and free end is moving with speed v 

Velocity at distance x from free end v = \frac{V}{L}x

 

Kinetic energy of mass dm at x = x is K.E = 1/2 (dm )\left [ \left ( \frac{V}{L} \right )^2 \right ]^L

 

\frac{1}{2}\left ( \frac{M}{L}dx \right )\frac{V^2}{L^2}x^2

\frac{M}{2L^3}v^2x^2 dx

Total kinetic energy = \int_{0}^{L}\frac{Mv^2}{2L^3}x^2dx = \frac{Mv^2}{2L^3}\cdot \frac{L^3}{3}= \frac{1}{6}Mv^2

 

 

 

 


Option 1)

\frac{1}{2}mv^{2}

Option 2)

mv^{2}

Option 3)

\frac{1}{3}mv^{2}

Option 4)

\frac{1}{6}mv^{2}

Posted by

SudhirSol

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