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Conservation of Energy

AP Physics 1· difficulty 4/5

A block slides along a frictionless surface at 5.0 m/s5.0\,\text{m/s}, crosses a rough patch of length 2.0 m2.0\,\text{m} with μk=0.40\mu_k=0.40, then continues onto frictionless ice. Take g=10 m/s2g=10\,\text{m/s}^2. What is its speed after leaving the rough patch?

  • A

    v≈3.0 m/sv\approx 3.0\,\text{m/s}

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  • B

    v≈2.0 m/sv\approx 2.0\,\text{m/s}

  • C

    v≈4.0 m/sv\approx 4.0\,\text{m/s}

  • D

    v≈5.0 m/sv\approx 5.0\,\text{m/s}

Explanation

Work-energy: 12vf2=12vi2−μkgd=12(25)−0.4(10)(2)=12.5−8=4.5\tfrac{1}{2}v_f^2=\tfrac{1}{2}v_i^2-\mu_k g d=\tfrac{1}{2}(25)-0.4(10)(2)=12.5-8=4.5. So vf=9=3.0 m/sv_f=\sqrt{9}=3.0\,\text{m/s}.

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