A 4.0 kg block sits at rest on a frictionless incline of angle \theta = 35°. What is the normal force on the block? Use g = 9.8\text{ m/s}^2.

32.1 N Perpendicular balance: N = mg \cos\theta = 4.0 \times 9.8 \times \cos 35° \approx 39.2 \times 0.8192 \approx 32.1\text{ N}.

Newton's Second Law (Translational Dynamics)

AP Physics 1· difficulty 4/5

\theta = 35°

Explanation

N = mg \cos\theta = 4.0 \times 9.8 \times \cos 35° \approx 39.2 \times 0.8192 \approx 32.1\text{ N}

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