JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 14)
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 :
_9th_January_Morning_Slot_en_14_1.png)
$${{2F} \over {\sqrt {mk} }}$$
$${F \over {\pi \sqrt {mk} }}$$
$${{\pi F} \over {\sqrt {mk} }}$$
$${F \over {\sqrt {mk} }}$$
Explanation
_9th_January_Morning_Slot_en_14_2.png)
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 = $${F \over K}$$
Applying work energy theorem, work done by all the forces = change in Kinetic energy.
$$ \Rightarrow $$ WForce + WSpring = $${1 \over 2}m{v^2} - 0$$
$$ \Rightarrow $$ Fx $$-$$ $${1 \over 2}$$Kx2 = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $$F$$$$\left( {{F \over K}} \right) - {1 \over 2}K{\left( {{F \over K}} \right)^2}$$ = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $${{{F^2}} \over K} - {{F{}^2} \over {2K}}$$ = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $${{F{}^2} \over {2K}}$$ = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$ $$v$$ = $${F \over {\sqrt {mK} }}$$
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