Question 151: Work, Energy and Power & Circular Motion

Question

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 :

Accepted Answer

When block is pulled x distance with F force, then it is distance with  F force,  then it is at maximum position
$$ 	herefore $$   F = Kx
$$ \Rightarrow $$   x = $${F \over K}$$
Applying work energy theorem, work done by all the forces = change in Kinetic energy.
$$ \Rightarrow $$  W
Force
+ W
Spring
= $${1 \over 2}m{v^2} - 0$$
$$ \Rightarrow $$   Fx $$-$$ $${1 \over 2}$$Kx
2
= $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$   $$F$$$$\left( {{F \over K}} ight) - {1 \over 2}K{\left( {{F \over K}} ight)^2}$$ = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$   $${{{F^2}} \over K} - {{F{}^2} \over {2K}}$$ = $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$   $${{F{}^2} \over {2K}}$$ =  $${1 \over 2}m{v^2}$$
$$ \Rightarrow $$   $$v$$ = $${F \over {\sqrt {mK} }}$$

Answer

$${{2F} \over {\sqrt {mk} }}$$

Answer

$${F \over {\pi \sqrt {mk} }}$$

Answer

$${{\pi F} \over {\sqrt {mk} }}$$

Answer

$${F \over {\sqrt {mk} }}$$

Work, Energy and Power & Circular Motion

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