Question

A spring of force constant K is first stretched by distance a from its natural length and then future by distance b. The work done in stretching the part b is

• A. $\frac{1}{2}$Ka(a-b)
• B. $\frac{1}{2}$Ka(a+b)
• C. $\frac{1}{2}$Kb(a-b)
• D. $\frac{1}{2}$Kb(2a+b)

Answer 1

The correct option is D $\frac{1}{2}$Kb(2a+b)

Work done by spring in its natural length$=\frac{1}{2}\times k\times x^2=\frac{1}{2}\times k\times a^2$

So, total work$=\frac{1}{2}k(a+b{)}^2$

for work done for stretching 'b'

$\frac{1}{2}\times k\times (a+b{)}^2-\frac{1}{2}\times k\times a^2=\frac{1}{2}\times k\times b\times (2a+b)$