Q. A spring of spring constant $5 \times 10^{3}Nm^{- 1}$ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is
A) 12.50 N-m
B) 18.75 N-m
C) 25.00 N-m
D) 6.25 N-m
Explanation
(b) $W_{1} = \frac{1}{2}kx_{1}^{2} = \frac{1 \times 5}{2} \times 10^{3} \times \left( 5 \times 10^{- 2} \right)^{2} = 6.25\ J$
$W_{2} = \frac{1}{2}k\left( x_{1} + x_{2} \right)^{2}$
$= \frac{1}{2} \times 5 \times 10^{3}\left( 5 \times 10^{- 2} + 5 \times 10^{- 2} \right)^{2} = 25\ J$
Net work done $= W_{2} - W_{1}$
$= 25 - 6.25 = 18.75\ J = 18.75\ N - m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.