Question

The potential energy of a certain spring when stretched through a distance x is 10 J. Th amount of work (in joules) that must be done on the spring to stretched it through an addition distance x will be

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The correct option is C 30

Initial PE to stretch it through a distance x is $U_i=\frac{1}{2}k(x{)}^2$

PE to stretch it through an additional distance x, So total distance x+x=$2$x

$U_f=\frac{1}{2}k(2x{)}^2$

$\therefore$ Work done $=\Delta U$

$=\frac{1}{2}k(2x{)}^2-\frac{1}{2}k(x{)}^2=3\left(\frac{1}{2}k{x}^2\right)$

$=3U_i=3\times 10=30J$.