Question

The P.E. of a certain spring when stretched from its natural length through a distance 0.3 m is10 J. The amount of work in joule that must be done on this spring to stretch it through an additional distance 0.15 m will be

• A. 10 J
• B. 20 J
• C. 7.5 J
• D. 12.5 J

Answer 1

The correct option is D

12.5 J

When x = 0.3m
$10=\frac{1}{2}k(0.3{)}^2$
When x = 0.45m
$U=\frac{1}{2}\times k\times (.45{)}^{2}U=\frac{1}{2}\times \frac{20}{(0.3{)}^2}\times (0.45{)}^2=10\left(\frac{3}{2}\right)$

= 22.5 J

$\therefore$ Additional energy is
U -10 =22.5-10

= 12.5J