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⚛️ Physics  ·  Laws of Motion  ·  NEET & JEE

A 1200 kg car moving at 20 m/s brakes to stop in 50 m. Braking force is:

Answer: 4800 N.

  • A 1200 N
  • B 2400 N
  • C 4800 N
  • D 9600 N

Correct answer: C. 4800 N

Explanation: v<sup>2</sup> = u<sup>2</sup> + 2as: 0 = 400 + 2a x 50, a = -4 m/s<sup>2.</sup> F = ma = 1200 x 4 = 4800 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 →