Q. 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 and comes to rest again after descending through a distance $x$.

Work, Energy and Power Report Error

A

$x=m g / k$

B

$x=2 m g / k$

C

The ball will have no acceleration at the position where it has descended through $x / 2$.

D

The ball will have an upward acceleration equal to $g$ at its lowermost position

Solution:

Apply conservation of energy:
loss in PE of ball = gain in PE of spring
$m g x=\frac{1}{2} k x^{2}$
$\Rightarrow x=2 m g / k$
At position $x / 2$ :

$k x / 2=m g$. So net force is zero. Hence, no acceleration at lowest position: $k x-m g=m a$
$\Rightarrow a=g$

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