Question

The spring is compressed by a distance of $a$ and released. The block again comes to rest when the spring is elongated by a distance $b$. During this process

• A. Work done by the spring on the block $=\frac{1}{2}k(a^2+b^2)$
• B. Work done by the spring on the block $=\frac{1}{2}k(a^2-b^2)$
• C. Coefficient of friction $=\frac{k(a-b)}{2\text{mg}}$
• D. Coefficient of friction $=\frac{k(a+b)}{2\text{mg}}$

Answer 1

Answer: (B), (C)

The correct options are
B Work done by the spring on the block $=\frac{1}{2}k(a^2-b^2)$
C Coefficient of friction $=\frac{k(a-b)}{2\text{mg}}$

Change in PE of spring
$=\frac{1}{2}k{b}^2-\frac{1}{2}k{a}^2=\frac{1}{2}k(b^2-a^2)$
Work done by spring =$-\Delta U=\frac{1}{2}k(a^2-b^2)$
Work done by friction $=-\mu \text{mg}(a+b)$
From Work energy theorem
$W_{friction}=\Delta K+\Delta U$
$W_{friction}=\Delta U$ $(\because \Delta K=0)$
$\frac{1}{2}k(b^2-a^2)=-\mu \text{mg}(a+b)$

$\therefore \mu =\frac{\frac{1}{2}k(a^2-b^2)}{\text{mg}(a+b)}=\frac{k}{2\text{mg}}(a-b)$