Question

A block suspended from a spring at natural length and is free to move vertically in the $y$-direction. Then

• A. Mass is released when $y=0$; the maximum value of $y$ reached by the mass $m$ is $2 m g / k$
• B. Mass is released when $y=0$; the maximum value of $y$ reached by the mass $m$ is $m g / k$
• C. Duc to air resistance the mass settles down into an equilibrium position $y_{\text {eq }}$, the mechanical energy lost is $\frac{1}{2}\left(m^{2} g^{2} / k\right)$
• D. Due to air resistance the mass settles down into an equilibrium position $y_{e q}$, the mechanical energy lost is $\left(m^{2} g^{2} / k\right)$

Answer 1

Answer: (A), (C)

For (1): $m g y=\frac{1}{2} k y^{2}$
$\Rightarrow y=\frac{2 m g}{k}$
Hence, (1) is correct, (2) is wrong.
For (3): $y_{e q}=\frac{m g}{k}$
Let velocity on reaching $y_{ eq }$ is $v$ when there was no air resistance.
Then
$m g y_{e q}=\frac{1}{2} k y_{e q}^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow m g\left(\frac{m g}{k}\right)=\frac{1}{2} k\left(\frac{m g}{k}\right)^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow \frac{1}{2} m v^{2}=\frac{m^{2} g^{2}}{2 k}$
Now this $KE$ will be lost when finally mass settles down at $y_{ eq }$ Hence, $(3)$ is correct, (4) is wrong.