Q. When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2.0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity of it in $ms^{- 1\ }is\ (g = 10\ ms^{- 2})$
A) 0.5
B) 1
C) 2
D) 4
Explanation
(b) $F = mg = kx;$
For first case,
$k = \frac{mg}{x} = \frac{1 \times 10N}{0.05\ m} = 200\ Nm^{- 1}$
For second case,
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10Hz$
$r = \frac{m^{'}g}{k} = \frac{2 \times 10}{200} = 0.1m$
$\therefore\ \ \ v_{\max} = r\omega = 0.1 \times 10 = 1\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.