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$A$ ball of mass $m$ is attached to the lower end of light vertical spring of force constant $k$. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length, comes to rest again after descending through a distance $x.$
The ball will have an upward acceleration equal to $g$ at its lowermost position.
$x = 2 mg/k$
The ball will have no acceleration at the position where it has descended through $x/2.$
All of the above
Solution
initial velocity $=$ final velocity $=0$ from energy conservation
$m g x-\frac{1}{2} k x^{2}=0$
$x=\frac{2 m g}{k}$
at descended length $=\frac{x}{2}$
$\frac{k x}{2}=k \cdot \frac{2 m g}{2 k}=m g$
Net force $=0$ $\Rightarrow a=0$ at lower most position.
force $=K x-m g=K \frac{2 m g}{K}-m g=m g$
$\Rightarrow a=g \uparrow$
