Question

A ball of mass $m$ is attached to the lower end of a 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 $x$. Given acceleration due to gravity $=g$.

• A. $x=\frac{mg}{k}$
• B. $x=\frac{2mg}{k}$
• C. The ball will have no acceleration at the position where it has descended through $\frac{x}{2}$.
• D. The ball will have an upward acceleration equal to $g$ at its lowermost position.

Answer 1

Answer: (B), (C), (D)

The correct options are
B The ball will have no acceleration at the position where it has descended through $\frac{x}{2}$.
C The ball will have an upward acceleration equal to $g$ at its lowermost position.
D $x=\frac{2mg}{k}$

The ball is at rest, it has maximum potential energy. When the ball is released from rest with the spring at its normal (unstretched) length it looses some potential energy and energy of spring increases. Hence, loss in potential potential energy of ball is equal to gain in potential potential energy of spring.
$\therefore mgx=\frac{1}{2}k{x}^2$
$\therefore x=\frac{2mg}{k}$
Also, for $x^{\prime }=\frac{x}{2}$,
$k{x}^{\prime }=mg$ i.e. forces are equal thus, the ball will have no acceleration at the position where it has descended through $\frac{x}{2}$.
And when ball is at lowermost position, the spring force will be
$kx=2mg$
Hence, the ball will have an upward acceleration equal to $g$ at its lowermost position.