Q. When a 4 kg mass is hung vertically on a light spring that obeys Hook’s law, the spring stretches by 2 cm. The work required to be done by an external agent in stretching this spring by 5 cm will be
A) 4.9 J
B) 2.45 J
C) 0.495 J
D) 0.245 J
Explanation
(b) $K\ = \frac{4\ \times 9.8}{2 \times 10^{- 2}}\ or\ \ K = 19.6\ \times 10^{2}\ Nm^{- 1}\ $
Work done = $\frac{1}{2} \times \ \ 19.6\ \times 10^{2} \times \left( 5 \times 10^{- 2} \right)^{2}\ J = 2.45\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.