AP Physics 1 · Topic 2.8

Spring Forces Practice

Part of Force and Translational Dynamics.(TOP-2.H)

Practice questions

6

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Sample questions

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  1. Sample 1difficulty 1/5

    A 10 kg10~\text{kg} object hangs from a vertical rope at rest. Using g10 m/s2g \approx 10~\text{m/s}^2, the rope tension is

    • A

      100 N100~\text{N}

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    • B

      00

    • C

      10 N10~\text{N}

    • D

      50 N50~\text{N}

    Why

    For equilibrium, T=mg=(10)(10)=100 NT = m g = (10)(10) = 100~\text{N}.

  2. Sample 2difficulty 2/5

    Two identical spring scales (each ideal) are connected end-to-end and a 10 kg10~\text{kg} mass is hung from the bottom. Using g10 m/s2g \approx 10~\text{m/s}^2, each scale reads

    • A

      200 N200~\text{N}

    • B

      Different values for each

    • C

      100 N100~\text{N}

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    • D

      50 N50~\text{N}

    Why

    Both scales are in series with the load and have the same tension everywhere along the rope: each reads W=100 NW = 100~\text{N}.

  3. Sample 3difficulty 2/5

    The cable of the elevator above snaps and the cabin falls freely. What does the scale now read?

    • A

      600 N600~\text{N}

    • B

      300 N300~\text{N}

    • C

      60 N60~\text{N}

    • D

      0 N0~\text{N}

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    Why

    In free fall a=ga = -g, giving N=m(g+a)=m(gg)=0 NN = m(g + a) = m(g - g) = 0~\text{N} — the familiar "weightlessness" result.

  4. Sample 4difficulty 2/5

    3 kg

    A 3 kg3~\text{kg} mass hangs in equilibrium from a spring with k=150 N/mk = 150~\text{N/m}. Using g10 m/s2g \approx 10~\text{m/s}^2, the stretch from natural length is

    • A

      0.15 m0.15~\text{m}

    • B

      0.05 m0.05~\text{m}

    • C

      0.20 m0.20~\text{m}

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    • D

      0.30 m0.30~\text{m}

    Why

    kx=mgx=30/150=0.20 mk x = m g \Rightarrow x = 30/150 = 0.20~\text{m}.

  5. Sample 5difficulty 3/5

    Two identical springs (each kk) connected in <strong>series</strong> behave as a single equivalent spring of constant

    • A

      4k4k

    • B

      2k2k

    • C

      k/2k/2

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    • D

      kk

    Why

    Total stretch under force FF: x=F/k+F/k=2F/kx = F/k + F/k = 2 F/k, so keq=F/x=k/2k_\text{eq} = F/x = k/2.