Work - AP Physics C Mechanics

  1. What is the work done by friction over a distance dd?
    W=−f×dW = -f \times d. Friction always opposes motion, doing negative work on moving objects.
  2. What happens to work done if force is halved, keeping displacement constant?
    Work is halved. Work is directly proportional to force in the formula.
  3. What is the work done by gravity on a free-falling object?
    W=m×g×hW = m \times g \times h. Gravitational work depends on mass, gravity, and vertical displacement.
  4. What happens to work done if displacement is doubled?
    Work doubles. Work is directly proportional to displacement in the formula.
  5. Identify the work-energy principle formula.
    KEf−KEi=W\text{KE}_f - \text{KE}_i = W. Net work equals the change in kinetic energy of an object.
  6. What is the formula for work when force and displacement are parallel?
    W=F×dW = F \times d. Work equals force times displacement when they're in the same direction.
  7. What is the condition for maximum work to be done by a force?
    Force is parallel to displacement. When θ=0°\theta = 0°, cos⁡(0°)=1\cos(0°) = 1, maximizing work done.
  8. What is the work done if a force of 10 N moves an object 5 m in the direction of the force?
    50 J50 \text{ J}. W=F×d=10×5=50W = F \times d = 10 \times 5 = 50 J when force is parallel.
  9. Which condition must be met for work to be zero?
    Force and displacement must be perpendicular. When θ=90°\theta = 90°, cos⁡(90°)=0\cos(90°) = 0, making work zero.
  10. What is the formula for power in terms of work?
    P=WtP = \frac{W}{t}. Power is the rate of doing work, measured in watts.
  11. What is work done by a force perpendicular to displacement?
    Zero. Perpendicular forces don't contribute to work in displacement direction.
  12. How is work affected by an increase in force?
    Work increases. Work is directly proportional to applied force magnitude.
  13. Identify the work done if F=20 NF = 20 \text{ N}, d=2 md = 2 \text{ m}, and θ=60°\theta = 60^\text{°}.
    20 J20 \text{ J}. W=20×2×cos⁡(60°)=20×2×0.5=20W = 20 \times 2 \times \cos(60°) = 20 \times 2 \times 0.5 = 20 J.
  14. What is work if F=18 NF = 18 \text{ N}, d=6 md = 6 \text{ m}, and θ=0°\theta = 0^\text{°}?
    108 J108 \text{ J}. W=18×6×cos⁡(0°)=18×6×1=108W = 18 \times 6 \times \cos(0°) = 18 \times 6 \times 1 = 108 J.
  15. Identify the angle used in the work formula W=F×d×cos(θ)W = F \times d \times \text{cos}(\theta).
    Angle between force and displacement. The angle θ\theta is measured between the force and displacement vectors.
  16. If a force is applied but no movement occurs, what is the work done?
    Zero. Work requires displacement; no movement means no work done.
  17. What is the work done if F=30 NF = 30 \text{ N}, d=5 md = 5 \text{ m}, and θ=180°\theta = 180^\text{°}?
    −150 J-150 \text{ J}. W=30×5×cos⁡(180°)=30×5×(−1)=−150W = 30 \times 5 \times \cos(180°) = 30 \times 5 \times (-1) = -150 J.
  18. Determine the work done when F=0 NF = 0 \text{ N}.
    Zero. No force means no work can be done regardless of displacement.
  19. Determine the change in work if the displacement is tripled.
    Work triples. Work is directly proportional to displacement in the formula.
  20. Find the work if F=15 NF = 15 \text{ N}, d=3 md = 3 \text{ m}, and θ=90°\theta = 90^\text{°}.
    0 J0 \text{ J}. W=15×3×cos⁡(90°)=15×3×0=0W = 15 \times 3 \times \cos(90°) = 15 \times 3 \times 0 = 0 J.
  21. How does work change if the angle between force and displacement is 90°90^\text{°}?
    Work is zero. At 90°, cosine equals zero, making the work calculation zero.
  22. Calculate the work done if F=10 NF = 10 \text{ N}, d=0 md = 0 \text{ m}, and θ=30°\theta = 30^\text{°}.
    0 J0 \text{ J}. W=10×0×cos⁡(30°)=0W = 10 \times 0 \times \cos(30°) = 0 J when displacement is zero.
  23. What is the direction of work done by friction?
    Opposite to displacement. Friction always acts opposite to the direction of relative motion.
  24. Identify the work done if F=8 NF = 8 \text{ N}, d=3 md = 3 \text{ m}, and θ=45°\theta = 45^\text{°}.
    16.97 J16.97 \text{ J}. W=8×3×cos⁡(45°)=8×3×0.707=16.97W = 8 \times 3 \times \cos(45°) = 8 \times 3 \times 0.707 = 16.97 J.
  25. What is the relationship between work and energy?
    Work is energy transfer. Work transfers energy from one form to another or between objects.
  26. Calculate the work done if F=20 NF = 20 \text{ N}, d=4 md = 4 \text{ m}, and θ=60°\theta = 60^\text{°}.
    40 J40 \text{ J}. W=20×4×cos⁡(60°)=20×4×0.5=40W = 20 \times 4 \times \cos(60°) = 20 \times 4 \times 0.5 = 40 J.
  27. What is the work done if F=5 NF = 5 \text{ N}, d=2 md = 2 \text{ m}, and θ=0°\theta = 0^\text{°}?
    10 J10 \text{ J}. W=5×2×cos⁡(0°)=5×2×1=10W = 5 \times 2 \times \cos(0°) = 5 \times 2 \times 1 = 10 J.
  28. What is the SI unit of work?
    Joule (J). Standard SI unit for energy and work in physics.
  29. Determine the work done if a force of 12 N moves an object 4 m at an angle of 90°90^\text{°}.
    0 J0 \text{ J}. W=12×4×cos⁡(90°)=12×4×0=0W = 12 \times 4 \times \cos(90°) = 12 \times 4 \times 0 = 0 J.
  30. What is the effect of increasing the angle θ\theta on work done?
    Work decreases. As θ\theta increases from 0°, cos⁡(θ)\cos(\theta) decreases, reducing work.
  31. Calculate the work done if F=25 NF = 25 \text{ N}, d=10 md = 10 \text{ m}, and θ=0°\theta = 0^\text{°}.
    250 J250 \text{ J}. W=25×10×cos⁡(0°)=25×10×1=250W = 25 \times 10 \times \cos(0°) = 25 \times 10 \times 1 = 250 J.
  32. State the formula for work done at an angle.
    W=F×d×cos(θ)W = F \times d \times \text{cos}(\theta). Cosine accounts for the angle between force and displacement vectors.
  33. Calculate the work done if F=10 NF = 10 \text{ N}, d=10 md = 10 \text{ m}, and θ=30°\theta = 30^\text{°}.
    86.6 J86.6 \text{ J}. W=10×10×cos⁡(30°)=10×10×0.866=86.6W = 10 \times 10 \times \cos(30°) = 10 \times 10 \times 0.866 = 86.6 J.
  34. What type of quantity is work: scalar or vector?
    Scalar. Work has magnitude but no direction, making it a scalar quantity.
  35. What does the term 'negative work' imply?
    Force opposes displacement. Force acts against motion direction, removing energy from the system.

Study these flashcards