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NEET PHYSICSEasy

A block of mass 2 kg2 \text{ kg} is placed on an inclined rough surface ACAC (as shown in the figure) of coefficient of friction μ\mu. If g=10 m s−2g=10 \text{ m s}^{-2}, the net force (in N) on the block will be:

A

10310\sqrt{3}

B

zero

C

1010

D

2020

Step-by-Step Solution

  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 θ\theta is less than or equal to the angle of repose (i.e., tan⁡θ≤μs\tan \theta \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 (Fnet=ma=0F_{net} = ma = 0).
  3. Forces Involved: The forces acting on the block are:
  • Weight (mgmg) vertically downwards.
  • Normal Reaction (NN) perpendicular to the incline.
  • Frictional Force (ff) along the incline opposing the tendency of motion.
  1. 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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