Zaymiey

⚛️ Physics  ·  Laws of Motion  ·  NEET & JEE

Block A (3 kg) on block B (5 kg) on frictionless floor. mu between A and B = 0.4. Max force on B for A not to slip (g=10):

Answer: 32 N.

  • A 12 N
  • B 32 N
  • C 40 N
  • D 50 N

Correct answer: B. 32 N

Explanation: Max friction on A: f = 0.4 x 3 x 10 = 12 N. Max a for A = 4 m/s<sup>2.</sup> Max F = (3+5) x 4 = 32 N.

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 →