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Kinetic and Static Friction

AP Physics 1· difficulty 2/5

A 5.0 kg5.0~\text{kg} block sits on a frictionless 37∘37^\circ incline (sin⁡37∘=0.60\sin 37^\circ = 0.60, cos⁡37∘=0.80\cos 37^\circ = 0.80). Using g≈10 m/s2g \approx 10~\text{m/s}^2, what is the normal force on the block?

  • A

    20 N20~\text{N}

  • B

    40 N40~\text{N}

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  • C

    50 N50~\text{N}

  • D

    30 N30~\text{N}

Explanation

N=mgcos⁡θ=(5.0)(10)(0.80)=40 NN = mg\cos\theta = (5.0)(10)(0.80) = 40~\text{N}.

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AP Physics 1 · Kinetic and Static Friction Practice Question | Acemy