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Conservation of Energy

AP Physics 1· difficulty 4/5

A skier of mass 60 kg60\,\text{kg} starts at rest from height h=20 mh=20\,\text{m} atop a curved hill and reaches the bottom moving at 14 m/s14\,\text{m/s}. Take g=10 m/s2g=10\,\text{m/s}^2. How much energy was dissipated by friction over the path?

  • A

    Eth≈1.20×104 JE_{th}\approx 1.20\times10^4\,\text{J} (initial PE, not dissipation)

  • B

    Eth≈5.88×103 JE_{th}\approx 5.88\times10^3\,\text{J} (final kinetic energy, not dissipation)

  • C

    Eth≈6.12×103 JE_{th}\approx 6.12\times10^3\,\text{J}

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  • D

    Eth=0E_{th}=0 (frictionless)

Explanation

Initial PE =mgh=60⋅10⋅20=12,000 J=mgh=60\cdot 10\cdot 20=12{,}000\,\text{J}. Final KE =12mv2=12(60)(142)=5,880 J=\tfrac{1}{2}mv^2=\tfrac{1}{2}(60)(14^2)=5{,}880\,\text{J}. Energy dissipated by friction =ΔE=12,000−5,880=6,120 J≈6.12×103 J= \Delta E = 12{,}000 - 5{,}880 = 6{,}120\,\text{J}\approx 6.12\times 10^3\,\text{J}.

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