AP Physics 1 · Topic 2.4

Newton's First Law (Translational Equilibrium) Practice

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

Practice questions

7

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Sample questions

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  1. Sample 1difficulty 1/5

    A book sits at rest on a table. Which statement correctly describes the forces on the book?

    • A

      The table pushes harder than gravity, but they balance.

    • B

      There are no forces on it because it's at rest.

    • C

      Only gravity acts on it.

    • D

      Gravity and the table's normal force act, summing to zero.

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    Why

    At rest, Fnet=0F_\text{net} = 0. Gravity (down) and normal force (up) are equal in magnitude and opposite in direction.

  2. Sample 2difficulty 2/5

    A 20 N20~\text{N} box rests on a floor with μs=0.40\mu_s = 0.40. A horizontal push of 5 N5~\text{N} is applied but the box doesn't move. What is the static friction on the box?

    • A

      8 N8~\text{N}

    • B

      5 N5~\text{N}

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    • C

      0 N0~\text{N}

    • D

      20 N20~\text{N}

    Why

    Static friction matches the applied force up to its maximum (μsN=0.40×20=8 N\mu_s N = 0.40 \times 20 = 8~\text{N}). Here applied is only 5 N5~\text{N}, so fs=5 Nf_s = 5~\text{N}.

  3. Sample 3difficulty 3/5

    A weight WW hangs from two ropes attached to a ceiling. Rope 1 makes 3030^\circ with vertical and rope 2 makes 6060^\circ with vertical. The tension T1T_1 in rope 1 (the more vertical one) is

    • A

      Larger than T2T_2.

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    • B

      Equal to T2T_2.

    • C

      Not enough info.

    • D

      Smaller than T2T_2.

    Why

    Vertical components must sum to WW: T1cos30+T2cos60=WT_1\cos 30 + T_2\cos 60 = W. Horizontal components cancel: T1sin30=T2sin60T_1\sin 30 = T_2\sin 60, giving T2=T1sin30/sin60=T1/3<T1T_2 = T_1\sin 30/\sin 60 = T_1/\sqrt 3 < T_1.

  4. Sample 4difficulty 3/5

    W T₁ (60° below horizontal) T₂ (horizontal)

    A weight W=50 NW = 50~\text{N} hangs in equilibrium. String 1 makes 6060^\circ below the horizontal ceiling and supports the weight. String 2 is horizontal. What is the tension in <strong>string 2</strong>?

    • A

      29 N29~\text{N}

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    • B

      25 N25~\text{N}

    • C

      50 N50~\text{N}

    • D

      58 N58~\text{N}

    Why

    Vertical: T1sin60=WT1=50/0.86658 NT_1 \sin 60^\circ = W \Rightarrow T_1 = 50/0.866 \approx 58~\text{N}. Horizontal: T2=T1cos6058×0.5=29 NT_2 = T_1 \cos 60^\circ \approx 58 \times 0.5 = 29~\text{N}.

  5. Sample 5difficulty 3/5

    A 1.0 kg1.0~\text{kg} and a 4.0 kg4.0~\text{kg} block connected by a string on a frictionless surface are pulled by a 25 N25~\text{N} force on the 4 kg4~\text{kg} block. What is the string tension?

    • A

      20 N20~\text{N}

    • B

      25 N25~\text{N}

    • C

      10 N10~\text{N}

    • D

      5 N5~\text{N}

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    Why

    System: a=25/5=5 m/s2a = 25/5 = 5~\text{m/s}^2. Isolate 1 kg1~\text{kg} block: T=ma=(1)(5)=5 NT = m a = (1)(5) = 5~\text{N}.