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Conservation of Energy

AP Physics 1· difficulty 3/5

A block slides 4 m4~\text{m} down a 30∘30^\circ incline starting from rest. Friction does −12 J-12~\text{J} of work on the 1.0 kg1.0~\text{kg} block (using g≈10 m/s2g \approx 10~\text{m/s}^2). The final speed at the bottom of the slide is

  • A

    2.0 m/s2.0~\text{m/s}

  • B

    2.8 m/s2.8~\text{m/s}

  • C

    5.7 m/s5.7~\text{m/s}

  • D

    4.0 m/s4.0~\text{m/s}

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Explanation

Vertical drop: h=4sin⁡30∘=2.0 mh = 4 \sin 30^\circ = 2.0~\text{m}. Energy at the bottom: Kf=mgh+Wfriction=(1.0)(10)(2.0)+(−12)=20−12=8 JK_f = mgh + W_{friction} = (1.0)(10)(2.0) + (-12) = 20 - 12 = 8~\text{J}. v=2Kf/m=2⋅8/1.0=16=4.0 m/sv = \sqrt{2K_f / m} = \sqrt{2 \cdot 8/1.0} = \sqrt{16} = 4.0~\text{m/s}.

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