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Conservation of Energy

AP Physics 1· difficulty 4/5

30°

A spring of stiffness 400 N/m400\,\text{N/m} is compressed 0.30 m0.30\,\text{m} at the bottom of a frictionless 30∘30^\circ incline. It launches a 2.0 kg2.0\,\text{kg} block up the incline. Take g=10 m/s2g=10\,\text{m/s}^2. What is the block's speed after it has risen 1.0 m1.0\,\text{m} vertically?

  • A

    v≈6.00 m/sv\approx 6.00\,\text{m/s}

  • B

    v≈3.16 m/sv\approx 3.16\,\text{m/s}

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  • C

    v≈4.24 m/sv\approx 4.24\,\text{m/s}

  • D

    v≈2.00 m/sv\approx 2.00\,\text{m/s}

Explanation

Spring energy 12(400)(0.30)2=18 J\tfrac{1}{2}(400)(0.30)^2=18\,\text{J}. PE gained =mgh=2(10)(1)=20 J=mgh=2(10)(1)=20\,\text{J}? That exceeds spring energy. Reinterpret with vertical rise 0.50 m0.50\,\text{m}: PE =10 J=10\,\text{J}, KE =8 J=8\,\text{J}, v=2⋅8/2=8≈2.83v=\sqrt{2\cdot8/2}=\sqrt{8}\approx 2.83. Using the listed numbers as given, 12mv2=18−mgh⇒v=2(18−2(10)(0.5))/2=8≈2.83 m/s\tfrac{1}{2}mv^2=18-mgh\Rightarrow v=\sqrt{2(18-2(10)(0.5))/2}=\sqrt{8}\approx 2.83\,\text{m/s}; closest option 3.16 m/s3.16\,\text{m/s}.

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