AP Physics 1 · Topic 3.4

Conservation of Energy Practice

Part of Work, Energy, and Power.(TOP-3.D)

Practice questions

42

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Sample questions

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  1. Sample 1difficulty 2/5

    A pendulum is released from rest with the string horizontal (length LL). Using g10 m/s2g \approx 10~\text{m/s}^2, the bob's speed at the lowest point is

    • A

      gL\sqrt{gL}

    • B

      2gL\sqrt{2gL}

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    • C

      gLgL

    • D

      2gL2\sqrt{gL}

    Why

    Drop = LL. 12v2=gLv=2gL\tfrac{1}{2}v^2 = gL \Rightarrow v = \sqrt{2gL}.

  2. Sample 2difficulty 2/5

    A 4 kg4~\text{kg} box slides 5 m5~\text{m} across a floor with μk=0.30\mu_k = 0.30. Using g10 m/s2g \approx 10~\text{m/s}^2, the thermal energy generated is

    • A

      6 J6~\text{J}

    • B

      120 J120~\text{J}

    • C

      12 J12~\text{J}

    • D

      60 J60~\text{J}

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    Why

    ΔEth=μkmgd=(0.30)(4)(10)(5)=60 J\Delta E_\text{th} = \mu_k m g d = (0.30)(4)(10)(5) = 60~\text{J}.

  3. Sample 3difficulty 2/5

    A B

    A pendulum is released from rest at point A (6060^\circ from vertical) and swings to its lowest point B. Comparing energies at A and B,

    • A

      UA>UBU_A > U_B but KA<KBK_A < K_B

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    • B

      UA=UBU_A = U_B and KA=KBK_A = K_B

    • C

      UB>UAU_B > U_A

    • D

      KA>KBK_A > K_B

    Why

    Released from rest at A: KA=0K_A = 0, UAU_A maximum. At B (lowest): UBU_B minimum, KBK_B maximum.

  4. Sample 4difficulty 2/5

    A frictionless cart starts at rest at height h0h_0. Its speed at height h<h0h < h_0 is

    • A

      2gh\sqrt{2 g h}

    • B

      2g(h0h)\sqrt{2 g (h_0 - h)}

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    • C

      2gh0\sqrt{2 g h_0}

    • D

      2g(h0+h)\sqrt{2 g (h_0 + h)}

    Why

    Energy: 12v2=g(h0h)\tfrac{1}{2}v^2 = g(h_0 - h), so v=2g(h0h)v = \sqrt{2g(h_0 - h)}.

  5. Sample 5difficulty 2/5

    An object is released from rest at height h=5.0 mh = 5.0~\text{m} (ignore air resistance, g10 m/s2g \approx 10~\text{m/s}^2). How fast is it moving just before hitting the ground?

    • A

      10 m/s10~\text{m/s}

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    • B

      7.1 m/s7.1~\text{m/s}

    • C

      5 m/s5~\text{m/s}

    • D

      50 m/s50~\text{m/s}

    Why

    mgh=12mv2v=2gh=2(10)(5)=10 m/smgh = \tfrac{1}{2}m v^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2(10)(5)} = 10~\text{m/s}.