Q. The force constant of a weightless spring is $16\ N/m$. A body of mass $1.0\ kg$ suspended from it is pulled down through 5 cm and then released. The maximum kinetic energy of the system (spring + body) will be
A) $2 \times 10^{- 2}\ J$
B) $4 \times 10^{- 2}\ J$
C) $8 \times 10^{- 2}\ J$
D) $16 \times 10^{- 2}\ J$
Explanation
(a) Max. K.E. of the system = Max. P.E. of the system
$\frac{1}{2}kx^{2} = \frac{1}{2} \times (16) \times \left( 5 \times 10^{- 2} \right)^{2} = 2 \times 10^{- 2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.