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Conservation of Energy

AP Physics 1· difficulty 4/5

A block of mass 2.0 kg2.0\,\text{kg} is released from rest at the top of a 30°30° incline of length 5.0 m5.0\,\text{m}. The coefficient of kinetic friction is μk=0.25\mu_k = 0.25. Use g=10 m/s2g = 10\,\text{m/s}^2. What is the block's speed at the bottom?

  • A

    approximately 5.7 m/s5.7\,\text{m/s}

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  • B

    approximately 4.0 m/s4.0\,\text{m/s}

  • C

    approximately 7.1 m/s7.1\,\text{m/s}

  • D

    approximately 10.0 m/s10.0\,\text{m/s}

Explanation

Height drop h=5sin⁡30°=2.5 mh = 5\sin 30° = 2.5\,\text{m} gives ΔU=mgh=50 J\Delta U = mgh = 50\,\text{J}. Friction loss is μkmgcos⁡θ⋅L=0.25⋅2⋅10⋅0.866⋅5≈21.7 J\mu_k mg\cos\theta \cdot L = 0.25\cdot 2\cdot 10\cdot 0.866\cdot 5 \approx 21.7\,\text{J}. Then 12mv2≈28.3 J\tfrac{1}{2}m v^2 \approx 28.3\,\text{J} gives v≈5.7 m/sv \approx 5.7\,\text{m/s}.

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