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⚛️ Physics  ·  Laws of Motion  ·  NEET & JEE

A 5 kg block rests on a rough surface with coefficient of friction 0.4. The minimum horizontal force needed to just start it moving is (g = 10 m s<sup>-2</sup>):

Answer: 20 N.

  • A 10 N
  • B 20 N
  • C 50 N
  • D 25 N

Correct answer: B. 20 N

Explanation: Limiting friction = &mu;mg = 0.4 × 5 × 10 = 20 N, which is the force needed to just overcome friction.

incline surfaceθmgNfmg sinθmg cosθ

Free-body diagram of a block on an incline: weight (mg) resolves into mg sinθ along the slope and mg cosθ into the slope, balanced by the normal force N and friction f.

Concept context

Newton's three laws, friction, circular motion, and free body diagrams.

Read the full Laws of Motion notes →