Q. To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is d. if the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is (force constant ,k)
A) $\frac{mg}{k}$
B) $2d$
C) $\frac{mg}{3k}$
D) 4d
Explanation
(b) To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is $x_{1}$ , then
Work done by weight of the block
=Potential energy stored in spring
$mg\ x = \frac{1}{2}kx^{2}$
$\therefore x = 2\frac{mg}{k}\ 2d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because d = \frac{mg}{k} \right)\ \ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.