Question

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

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

Answer 1

Answer: (D)

$W_{1}=\frac{1}{2} k x^{2}=\frac{1}{2} k a^{2}$
$W_{2}=\frac{1}{2} k(a+b)^{2}$
$\Delta W=W_{2}-W_{1}=\frac{1}{2} k b(2 a+b)$