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AP Physics 1 · guided topic map

Work, energy and power for AP Physics 1

Work, energy and power for AP Physics 1, organized into 1 syllabus topic and 6 mapped concept guides.

Syllabus topics
1
Mapped concept guides
6
Educational level
AP Physics 1: Algebra-Based

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Work in order or jump to the concept named in your specification, course outline, or assignment.

3

Work, Energy, and Power

AP Physics 1: Algebra-Based

6 guides
  1. 01Work done by a force18–23%
  2. 02Force-displacement graphs18–23%
  3. 03Kinetic and potential energy18–23%
  4. 04Energy conversion and conservation18–23%
  5. 05The work-energy theorem18–23%
  6. 06Power and efficiency18–23%

Diagrams

Work, energy and power as AP Physics 1 draws it

The figures from the AP Physics 1 practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 7.1Force and Translational Dynamics · Work, Energy, and PowerAP
Block on a rough incline of angle θ leading onto a rough horizontal surfacemhθμμreleased from rest

Figure comment

Fig. 7.1A wedge-shaped incline stands on a horizontal floor with its sloping face rising from right to left, and the angle between the sloping face and the floor at the foot of the slope is marked θ. A block labelled m rests on the sloping face near the top and is noted as released from rest, with a dimension line to the left of the wedge marking its height h above the floor. The symbol μ is printed on the wedge below the sloping face and again on the floor beyond the foot of the slope, showing that the same coefficient of kinetic friction applies to both surfaces, which run into one another at the bottom of the incline.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

2 Force and Translational Dynamics3 Work, Energy, and PowerOpen AP question · AP Physics 1: Algebra-Based Q7
Guided questions 5 parts

Reading cue. h is the vertical dimension line at the side of the wedge, not the distance the block slides: along the face that is h/sin θ. μ is printed twice because both surfaces share it.

  1. aDetermine Determine the normal force exerted on the block by the sloping face, in terms of m, θ and physical constants.

    recall2 marks

    Check answer 2 marks
    1. resolves the weight mg perpendicular to the sloping face
    2. N = mg cos θ, since the block has no acceleration perpendicular to the face

  2. bCalculate Calculate the acceleration of the block down the sloping face for θ = 30° and μ = 0.25.

    routine3 marks

    Check answer 3 marks
    1. along the face, ma = mg sin θ − μmg cos θ, so a = g(sin θ − μ cos θ) with m cancelling
    2. a = 9.8(0.500 − 0.25 × 0.866) = 9.8 × 0.284
    3. a = 2.8 m/s², directed down the slope

  3. cDetermine Determine the time the block takes to reach the foot of the slope when h = 1.5 m, for the same θ and μ.

    demanding4 marks

    Check answer 4 marks
    1. the dimension line gives the vertical drop, so the distance along the face is L = h/sin θ = 1.5/0.500 = 3.0 m
    2. from rest with uniform acceleration, L = ½at², so t = √(2L/a)
    3. t = √(2 × 3.0/2.78) = √2.16
    4. t = 1.5 s

  4. dJustify An identical block is released from the same height h on a steeper wedge carrying the same μ on both of its surfaces. Justify whether it stops nearer to or further from the foot of the slope than the block in the figure.

    top of the paper4 marks

    Check answer 4 marks
    1. the friction force on the face is μmg cos θ and the sliding length is h/sin θ, so the energy lost on the slope is μmgh cot θ
    2. cot θ falls as θ increases, so the steeper wedge takes less energy from the block
    3. the block therefore reaches the foot with more kinetic energy, while the friction force on the floor is μmg and is unchanged by the wedge
    4. so it stops further from the foot of the slope, not nearer

Transfer challenge

A crate is given a push and slides 6.0 m up a ramp inclined at 20° before stopping. The coefficient of kinetic friction between crate and ramp is 0.30. Determine the speed of the crate at the start of the slide, and determine whether it then slides back down.

Check answer 4 marks
  1. moving up the slope, gravity and friction both act down it, so a = g(sin θ + μ cos θ)
  2. a = 9.8(0.342 + 0.30 × 0.940) = 6.1 m/s²
  3. v² = 2 × 6.1 × 6.0 = 73, so v = 8.6 m/s
  4. tan 20° = 0.36 exceeds μ = 0.30, so the component of weight along the slope beats the maximum friction and the crate slides back down
02Fig. 10.1Work, Energy, and Power · Torque and Rotational DynamicsAP
A sphere and a block released from the same height on two sections of one inclinetwo sections of the same inclineboth released from rest at the same heighthMMsphere rolls without slippingblock on a frictionless section

Figure comment

Fig. 10.1Two identical wedge-shaped inclines stand side by side on the same horizontal floor, labelled as two sections of the same incline. A solid sphere of mass M rests on the sloping face of the left wedge, and a block of mass M rests at the same point up the sloping face of the right wedge. A dashed horizontal line runs across the figure at the level of both objects, and a dimension line at the far left marks their common release height h above the floor. A note states that both are released from rest at the same height; a label under the left ramp reads that the sphere rolls without slipping, and one under the right ramp that the block is on a frictionless section.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

3 Work, Energy, and Power5 Torque and Rotational Dynamics6 Energy and Momentum of Rotating SystemsOpen AP question · AP Physics 1: Algebra-Based Q10
Guided questions 5 parts

Reading cue. The dashed line exists to fix that both start at the same height h; the labels under the two ramps — rolls without slipping, frictionless — are the whole difference between the cases.

  1. aIndicate Indicate whether the gravitational potential energy converted by the sphere on its way to the floor is greater than, less than, or equal to that converted by the block. No justification is required.

    recall1 mark

    Check answer 1 mark
    1. equal: both have mass M and both fall through the same height h from the dashed line, so each converts Mgh

  2. bDetermine Determine what fraction of the sphere's total kinetic energy at the foot of the ramp is rotational.

    routine3 marks

    Check answer 3 marks
    1. rolling without slipping gives ω = v/R, so the rotational term is ½(2/5)MR²(v/R)² = (1/5)Mv²
    2. total kinetic energy = ½Mv² + (1/5)Mv² = (7/10)Mv²
    3. fraction rotational = (1/5)/(7/10) = 2/7, about 0.29

  3. cDetermine Both objects cover the same distance along the face of their identical ramps, starting from rest with uniform acceleration. Determine the ratio of the time the sphere takes to reach the floor to the time the block takes.

    demanding4 marks

    Check answer 4 marks
    1. for uniform acceleration from rest the average speed is half the final speed, so the face length L gives t = 2L/v for each object
    2. L is the same for both, so the ratio of times is the inverse ratio of the final speeds: t_sphere/t_block = v_block/v_sphere
    3. v_block = √(2gh) and v_sphere = √(10gh/7), so the ratio is √(2 ÷ 10/7) = √(7/5)
    4. t_sphere/t_block = √1.4 = 1.18, so the sphere takes about 18% longer

  4. dDerive The sphere on the left ramp is replaced by a hollow spherical shell of the same mass and radius, for which I = (2/3)MR², released from rest on the same dashed line. Derive its speed at the floor and explain where it ranks against the two speeds the figure compares.

    top of the paper4 marks

    Check answer 4 marks
    1. energy conservation with ω = v/R: Mgh = ½Mv² + ½(2/3)MR²(v/R)² = (5/6)Mv²
    2. v_shell = √(6gh/5) = 1.10√(gh), against 1.20√(gh) for the solid sphere and 1.41√(gh) for the block
    3. the shell is slowest: v_shell/v_sphere = √(0.84) = 0.92, so it is about 8% slower than the sphere
    4. all of the shell's mass sits at the rim, giving the largest moment of inertia for the same M and R, so the largest share of the Mgh goes into rotation and the least into translation

Transfer challenge

A solid cylinder, for which I = ½MR², rolls without slipping along a horizontal floor at 3.0 m/s and then rolls up a ramp. Determine the vertical height it reaches, and compare it with the height a frictionless sliding block of the same mass and speed would reach.

Check answer 4 marks
  1. rolling gives ω = v/R, so the total kinetic energy is ½Mv² + ½(½MR²)(v/R)² = ¾Mv²
  2. at the highest point all of it has become Mgh, so h = 3v²/(4g)
  3. h = 3(3.0)²/(4 × 9.8) = 27/39.2 = 0.69 m
  4. a block sliding up a frictionless ramp at 3.0 m/s reaches only v²/2g = 0.46 m, because it carries no rotational kinetic energy to convert