⚛️ Physics · Laws of Motion · NEET & JEE

A 10 kg block rests on a floor with μ = 0.5. A horizontal force of 40 N is applied (g = 10 m/s²). The friction force acting on the block is:

Answer: 40 N.

A 50 N
B 40 N
C 20 N
D 0 N

Correct answer: B. 40 N

Explanation: Maximum static friction = 0.5·10·10 = 50 N > 40 N, so block stays still and friction = applied force = 40 N.

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.

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A 10 kg block rests on a floor with μ = 0.5. A horizontal force of 40 N is applied (g = 10 m/s²). The friction force acting on the block is:

Concept context

More Laws of Motion questions