Q.
A block of mass $m = 25$kg sliding on a smooth horizontal surface with a velocity $v = 3{ms}^{- 1}$ meets the spring of spring constant $k = 100{Nm}^{- 1}$ fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are
A) $1.5\ m,\ - 3\ {ms}^{- 1}$
B) $1.5\ m,\ 0.01\ {ms}^{- 1}$
C) $1.0\ m,\ 3\ {ms}^{- 1}$
D) $0.5\ m,\ 2\ {ms}^{- 1}$
Explanation
(a) When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
$\frac{1}{2}mv^{2} = \frac{1}{2}kx^{2}$
Or $x = \sqrt{\frac{mv^{2}}{k}}$
$= \sqrt{\frac{25{\times 3}^{2}}{100}}\ \ \sqrt{\frac{9}{4}}\ = \frac{3}{2} = 1.5\ m$
When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
$Hence\ v = - 3ms^{- 1}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.