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Newton's Second Law (Translational Dynamics)

AP Physics 1· difficulty 3/5

A 1.0 kg1.0~\text{kg} block is pushed at constant velocity across a rough floor by a horizontal force. If μk=0.30\mu_k = 0.30, the applied force is (g≈10 m/s2g \approx 10~\text{m/s}^2):

  • A

    30 N30~\text{N}

  • B

    3.0 N3.0~\text{N}

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  • C

    0.3 N0.3~\text{N}

  • D

    10 N10~\text{N}

Explanation

Constant velocity → applied = friction = μkmg=0.30⋅1⋅10=3.0 N\mu_k m g = 0.30 \cdot 1 \cdot 10 = 3.0~\text{N}.

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