A block of mass $2 ext{ kg}$ is placed on an inclined rough surface $AC$ (as shown in the figure) of coeffic

Question

A block of mass $2 	ext{ kg}$ is placed on an inclined rough surface $AC$ (as shown in the figure) of coefficient of friction $\mu$. If $g=10 	ext{ m s}^{-2}$, the net force (in N) on the block will be:

Accepted Answer

1. **Analyze the State of Motion:** The question asks for the 'net force' on a block placed on a rough incline. The probable answer 'zero' implies that the block is in a state of **equilibrium** (static equilibrium). This occurs when the angle of inclination $	heta$ is less than or equal to the angle of repose (i.e., $	an 	heta \le \mu_s$).
2. **Apply Newton's First Law:** According to Newton's First Law of Motion, if a body is at rest or moving with constant velocity, the vector sum of all external forces acting on it is zero ($F_{net} = ma = 0$).
3. **Forces Involved:** The forces acting on the block are:
   - Weight ($mg$) vertically downwards.
   - Normal Reaction ($N$) perpendicular to the incline.
   - Frictional Force ($f$) along the incline opposing the tendency of motion.
4. **Conclusion:** Since the block is stationary (held by friction), these three forces cancel each other out completely. Therefore, the net force on the block is **zero**.
   (Reference: NCERT Class 11, Physics Part I, Chapter 5: Laws of Motion, Section 5.8 discusses Equilibrium of a particle and Section 5.9 discusses Friction).

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