Q. A block whose mass is 650 g is fastened to a spring whose spring constantly is $65\ Nm^{- 1}.$ The block is pulled a distance $x = 11$ cm from its equilibrium position at $x = 0.$ On a frictionless surface and released from rest at $t = 0.$The maximum velocity of the vibrating block is
A) $1.1\ ms^{- 1}$
B) $0.65\ ms^{- 1}$
C) $1.30\ ms^{- 1}$
D) $2.6\ ms^{- 1}$
Explanation
(a) Max, KE= Max. PE
$= \frac{1}{2}mv^{2} = \frac{1}{2}kx^{2} = \frac{1}{2} \times 65 \times (0.11)^{2}$
or $v^{2} = \frac{65 \times (0.11)^{2}}{650 \times 10^{- 3}}$ or $v = 1.1\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.