Friction Basics question from JEE Main 2013 (25 Apr, Shift 1)

Answer: (b)

1. While $kt \le \mu_s mg$, static friction balances $F$ and $a = 0$.
2. Once the block starts, friction drops to $\mu_k mg < \mu_s mg$, so $a$ jumps from zero to $(\mu_s - \mu_k)g$.
3. After that $a = \dfrac{kt}{m} - \mu_k g$, a straight line with slope $\dfrac{k}{m}$.
4. Graph: zero, then a sudden jump, then a rising line, so the answer is (b).