AP Physics 1 · Topic 3.2

Work and Mechanical Energy Practice

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

Practice questions

15

Want a predicted score for the whole AP AP1 exam? Take the 20-question diagnostic and Lumi will plan the rest.

Sample questions

5 of 15 — sign in to practice the rest with adaptive difficulty and mastery tracking.

  1. Sample 1difficulty 1/5

    A satellite in circular orbit experiences gravity perpendicular to its motion. The work done by gravity in one full orbit is

    • A

      Positive

    • B

      Cannot be determined

    • C

      Zero

      check_circle
    • D

      Negative

    Why

    W=Fdcos90=0W = Fd\cos 90^\circ = 0. Gravity always points toward the center, perpendicular to velocity, so it does no work over the orbit.

  2. Sample 2difficulty 1/5

    Friction between a sliding block and the floor does work on the block. The sign of this work is

    • A

      Negative

      check_circle
    • B

      Cannot be determined

    • C

      Positive

    • D

      Zero

    Why

    Friction opposes motion; the angle is 180180^\circ, so W<0W < 0.

  3. Sample 3difficulty 1/5

    A 30 N30~\text{N} horizontal force pushes a box 4.0 m4.0~\text{m} across a floor in the direction of the force. How much work does the force do?

    • A

      30 J30~\text{J}

    • B

      60 J60~\text{J}

    • C

      7.5 J7.5~\text{J}

    • D

      120 J120~\text{J}

      check_circle

    Why

    W=Fdcosθ=(30)(4.0)(cos0)=120 JW = F d \cos\theta = (30)(4.0)(\cos 0^\circ) = 120~\text{J}.

  4. Sample 4difficulty 2/5

    0 x stretch F

    For a spring obeying Hooke's law, the elastic PE stored when stretched by xx equals

    • A

      2Fx2 F x

    • B

      12Fx\tfrac{1}{2} F x (triangle area)

      check_circle
    • C

      Fx\sqrt{F x}

    • D

      FxF x (rectangle area)

    Why

    PE = area under FF-vs-xx for a spring = triangle of base xx and height F=kxF = k x, so Us=12kx2=12FxU_s = \tfrac{1}{2} k x^2 = \tfrac{1}{2} F x.

  5. Sample 5difficulty 2/5

    0 4 6 x (m) F (N)

    The constant force shown acts over 00 to 4 m4~\text{m}. The work done is

    • A

      10 J10~\text{J}

    • B

      24 J24~\text{J}

      check_circle
    • C

      1.5 J1.5~\text{J}

    • D

      48 J48~\text{J}

    Why

    Work = area under FF-vs-xx: 6×4=24 J6 \times 4 = 24~\text{J}.