Q. A spring, which is initially in its unstretched condition, is first stretched by a length$x$ and then again by a further length $x$.The work done in the first case is $w_{1},\ $and in the second case is $w_{2}.Then$
A) $W_{2} = W_{1}$
B) $W_{2} = 2W_{1}$
C) $w_{2} = 3w_{1}$
D) $w_{2} = 4w_{1}$ $\ \ $
Explanation
(c) The work done in stretching a sprig by a length $x,$
${\ W}_{1} = \frac{1}{2}kx^{2}$ …(i)
The work done in stretching the spring by a further length$x.$
${\ \ \ \ \ \ W}_{2} = \frac{1}{2}k(2x)^{2} - \frac{1}{2}kx^{2}$
Or ${\ \ \ \ W}_{2} = \frac{1}{2}k \times 4x^{2} - \frac{1}{2}kx^{2}$
Or ${\ \ W}_{2} = 3 \times \frac{1}{2}kx^{2}$ ...(ii)
From Esq. (i) and (ii)we have
$W_{2} = 3W_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.