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QJEEM19Ph815b: JEE Main 2019 Physics - Class 11 JEEM19Ph815b Solution: A block of mass m, lying on a smooth horizontal surface, is…

A block of mass m, lying on a smooth horizontal surface, is attached to a sring (of negligible mass) of spring constant k. The other end of the 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 :
JEE Main 2019 (Online) 9th January Morning Slot Physics - Work Power & Energy Question 102 English

Multiple Choice Options

  • A 2Fmk{{2F} \over {\sqrt {mk} }}
  • B Fπmk{F \over {\pi \sqrt {mk} }}
  • C πFmk{{\pi F} \over {\sqrt {mk} }}
  • D Fmk{F \over {\sqrt {mk} }}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Morning Slot Physics - Work Power & Energy Question 102 English Explanation
When block is pulled x distance with F force, then it is distance with F force, then it is at maximum position

∴\therefore    F = Kx

⇒\Rightarrow    x = FK{F \over K}

Applying work energy theorem, work done by all the forces = change in Kinetic energy.

⇒\Rightarrow   WForce + WSpring = 12mv2−0{1 \over 2}m{v^2} - 0

⇒\Rightarrow    Fx −- 12{1 \over 2} Kx2 = 12mv2{1 \over 2}m{v^2}

⇒\Rightarrow     FF (FK)−12K(FK)2\left( {{F \over K}} \right) - {1 \over 2}K{\left( {{F \over K}} \right)^2} = 12mv2{1 \over 2}m{v^2}

⇒\Rightarrow     F2K−F22K{{{F^2}} \over K} - {{F{}^2} \over {2K}} = 12mv2{1 \over 2}m{v^2}

⇒\Rightarrow     F22K{{F{}^2} \over {2K}} = 12mv2{1 \over 2}m{v^2}

⇒\Rightarrow     vv = FmK{F \over {\sqrt {mK} }}

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