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Conservation of Energy

AP Physics 1· difficulty 3/5

A 2.0 kg2.0~\text{kg} block slides 3.0 m3.0~\text{m} down a 30∘30^\circ incline with μk=0.10\mu_k = 0.10. Using g≈10 m/s2g \approx 10~\text{m/s}^2 and starting from rest, what is the block's speed at the bottom?

  • A

    7.7 m/s7.7~\text{m/s}

  • B

    5.0 m/s5.0~\text{m/s}

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  • C

    3.8 m/s3.8~\text{m/s}

  • D

    5.5 m/s5.5~\text{m/s}

Explanation

Drop: h=(3.0)sin⁡30∘=1.5 mh = (3.0)\sin 30^\circ = 1.5~\text{m}, so ΔU=mgh=(2)(10)(1.5)=30 J\Delta U = mgh = (2)(10)(1.5) = 30~\text{J}. Friction loss: Wf=μkmgcos⁡θ⋅d=(0.10)(2)(10)(cos⁡30∘)(3.0)≈5.2 JW_f = \mu_k mg\cos\theta \cdot d = (0.10)(2)(10)(\cos 30^\circ)(3.0) \approx 5.2~\text{J}. Final K=30−5.2=24.8 JK = 30 - 5.2 = 24.8~\text{J}, so v=2K/m=24.8≈5.0 m/sv = \sqrt{2K/m} = \sqrt{24.8} \approx 5.0~\text{m/s}.

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