Newton's Second Law (Translational Dynamics)

AP Physics 1· difficulty 1/5

A 3.0 kg3.0~\text{kg} object's velocity goes from 4 m/s4~\text{m/s} east to 10 m/s10~\text{m/s} east in 3.0 s3.0~\text{s}. The net force on it is

  • A

    6 N6~\text{N}

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  • B

    2 N2~\text{N}

  • C

    10 N10~\text{N}

  • D

    14 N14~\text{N}

Explanation

a=(104)/3=2 m/s2a = (10-4)/3 = 2~\text{m/s}^2. F=ma=6 NF = m a = 6~\text{N}.

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