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  • Can someone explain A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulle

 

A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is: 

  • Option 1)

      \frac{2F}{\sqrt{mk}}

      

      

     

  • Option 2)

    \frac{F}{\pi \sqrt{mk}}

  • Option 3)

    \frac{\pi F}{\sqrt{mk}}

  • Option 4)

    \frac{F}{\sqrt{mk}}

Answers (1)

best_answer

 

Net work done by all the forces give the change in kinetic energy -

W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}

W= k_{f}-k_{i}

- wherein

m=mass \: of\: the\: body

v_{0}= initial\: velocity

v= final\: velocity

 

 

Let the block is pulled to a distance x

F = kx

x = \frac{F}{k} \ \ - (1)

Applying work energy theorem

Work done due to F + Work done by spring = Change in kinetic energy

i.e F.x - \frac{1}{2} kx2 = \frac{1}{2} mv2 - 0      (2)

Here work done due to spring is taken as negative because it is against the force applied.

From (1) and (2)

F.\frac{F}{K} - \frac{1}{2} K \left ( \frac{F}{K^{2}} \right ) = \frac{1}{2} m v^{2}

\frac{F^{2}}{K} - \frac{F^{2}}{2K} = \frac{1}{2} m v^{2}

v = \frac{F}{\sqrt{mK}}

 


Option 1)

  \frac{2F}{\sqrt{mk}}

  

  

 

Option 2)

\frac{F}{\pi \sqrt{mk}}

Option 3)

\frac{\pi F}{\sqrt{mk}}

Option 4)

\frac{F}{\sqrt{mk}}
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