A block of mass 4 kg is on a rough incline of 37 (sin37 = 0.6, cos37 = 0.8) with = 0.5. Its acceleration down the incline is (g = 10 m s -2 ):

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A block of mass 4 kg is on a rough incline of 37° (sin37° = 0.6, cos37° = 0.8) with μ = 0.5. Its acceleration down the incline is (g = 10 m s^-2):

Answer: 2 m s^-2.

A 3 m s^-2
B 1 m s^-2
C 2 m s^-2
D 4 m s^-2

Correct answer: C. 2 m s^-2

Explanation: a = g(sinθ - μcosθ) = 10(0.6 - 0.5 × 0.8) = 10(0.6 - 0.4) = 2 m s^-2.

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 block of mass 4 kg is on a rough incline of 37&deg; (sin37&deg; = 0.6, cos37&deg; = 0.8) with &mu; = 0.5. Its acceleration down the incline is (g = 10 m s-2):

Concept context

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A block of mass 4 kg is on a rough incline of 37° (sin37° = 0.6, cos37° = 0.8) with μ = 0.5. Its acceleration down the incline is (g = 10 m s^-2):