Given:
Mass \(m=0.5\ \text{kg}\), spring constant \(k=2\ \text{N\,m}^{-1}\), damping (drag) coefficient \(b=3\ \text{kg\,s}^{-1}\).
For critical damping:
The equation is \(m_{\rm total}\,x'' + b\,x' + kx = 0\). Critical damping occurs when \[ b = 2\sqrt{k\,m_{\rm total}}. \] Solve for the required total mass \(m_{\rm total}\): \[ m_{\rm total}=\frac{b^2}{4k}. \] Compute \(m_{\rm total}\): \[ m_{\rm total}=\frac{3^2}{4\times 2}=\frac{9}{8}=1.125\ \text{kg}. \] Additional mass required:
\[ \Delta m = m_{\rm total}-m = 1.125-0.5 = 0.625\ \text{kg}. \] Final answer (rounded to three decimals):
\[ \boxed{\Delta m = 0.625\ \text{kg}} \]
In order to achieve the static equilibrium of the see-saw about the fulcrum \( P \), shown in the figure, the weight of Box B should be _________ kg, if the weight of Box A is 50 kg.

A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{π}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer) 
(Take g = 10m/s2)

