AP Physics 1 Flashcards: Work

What this deck covers

This deck focuses on Work, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 1.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the work done if $F = 0$?

Answer: Zero, as no force means no work is done. Work requires both force and displacement to be nonzero.

Flashcard 2: Calculate work: $F = 30 \text{ N}$, $d = 2 \text{ m}$, $\theta = 0^\text{o}$.

Answer: 60 J. Using $W = Fd\cos(0°) = 30 × 2 × 1 = 60$ J.

Flashcard 3: What is the work done by a force in the direction opposite to displacement?

Answer: Negative work. When $\theta = 180°$, the cosine equals $-1$.

Flashcard 4: What does it mean if net work is zero?

Answer: The object's kinetic energy remains constant. No net energy transfer occurs to the object.

Flashcard 5: What is the work done by a conservative force over a closed path?

Answer: Zero. Conservative forces store and return energy without loss.

Flashcard 6: Find force $F$: $W = 60 \text{ J}$, $d = 4 \text{ m}$, $\theta = 0^\text{o}$.

Answer: $F = 15 \text{ N}$. From $W = Fd$, so $F = W/d = 60/4 = 15$ N.

Flashcard 7: What is the work done when no displacement occurs?

Answer: Zero, as work requires displacement. Work formula requires movement to occur.

Flashcard 8: Calculate work: $F = 50 \text{ N}$, $d = 2 \text{ m}$, $\theta = 30^\text{o}$.

Answer: 86.6 J. Using $\cos(30°) = \frac{\sqrt{3}}{2} ≈ 0.866$, so $W = 50 × 2 × 0.866 = 86.6$ J.

Flashcard 9: Find displacement $d$: $W = 50 \text{ J}$, $F = 10 \text{ N}$, $\theta = 0^\text{o}$.

Answer: $d = 5 \text{ m}$. Rearranging gives $d = W/F = 50/10 = 5$ m.

Flashcard 10: What is the relation between work and power?

Answer: Power is the rate at which work is done. Power equals work divided by time taken.

Flashcard 11: Calculate work: $F = 25 \text{ N}$, $d = 3 \text{ m}$, $\theta = 60^\text{o}$.

Answer: 37.5 J. Using $\cos(60°) = 0.5$, so $W = 25 × 3 × 0.5 = 37.5$ J.

Flashcard 12: What happens to work when force is doubled?

Answer: Work doubles, assuming displacement and angle are constant. Work is directly proportional to applied force magnitude.

Flashcard 13: Find displacement: $W = 200 \text{ J}$, $F = 50 \text{ N}$, $\theta = 0^\text{o}$.

Answer: $d = 4 \text{ m}$. From $W = Fd$, so $d = W/F = 200/50 = 4$ m.

Flashcard 14: Define negative work.

Answer: Work is negative when force opposes displacement. Force component removes energy from the system.

Flashcard 15: Identify the scalar quantity: work or force?

Answer: Work is a scalar quantity. Work has magnitude only, no direction like force vectors.

Flashcard 16: State the SI unit of work.

Answer: Joule (J). Same as Newton-meter (N⋅m) in physics.

Flashcard 17: Find work done: $F = 10 \text{ N}$, $d = 5 \text{ m}$, $\theta = 0^\text{o}$.

Answer: 50 J. Using $W = Fd\cos(0°) = 10 × 5 × 1 = 50$ J.

Flashcard 18: If $W = 100 \text{ J}$ and $d = 5 \text{ m}$, find force $F$ for $\theta = 0^\text{o}$.

Answer: $F = 20 \text{ N}$. Rearranging $W = Fd$ gives $F = W/d = 100/5 = 20$ N.

Flashcard 19: Calculate power: $W = 120 \text{ J}$, time $t = 4 \text{ s}$.

Answer: 30 W. Using $P = W/t = 120/4 = 30$ W.

Flashcard 20: Define positive work.

Answer: Work is positive when force and displacement are in the same direction. Force component aids the motion, adding energy to the system.

Flashcard 21: What is an example of a non-conservative force?

Answer: Friction. Force depends on the path taken between points.

Flashcard 22: What is the work done if the angle $\theta$ is $90^\text{o}$?

Answer: Zero, as $\text{cos}(90^\text{o}) = 0$. No force component exists parallel to displacement direction.

Flashcard 23: State the work-energy theorem.

Answer: The work done equals the change in kinetic energy. Fundamental principle connecting force and energy in mechanics.

Flashcard 24: Calculate work: $F = 20 \text{ N}$, $d = 3 \text{ m}$, $\theta = 90^\text{o}$.

Answer: 0 J. Since $\cos(90°) = 0$, making $W = 20 × 3 × 0 = 0$ J.

Flashcard 25: What is the work-energy principle?

Answer: Net work done on an object equals its change in kinetic energy. Also known as the work-energy theorem in physics.

Flashcard 26: How is work calculated when force is perpendicular to displacement?

Answer: Work is zero because $\text{cos}(90^\text{o}) = 0$. Perpendicular force components do no work on the object.

Flashcard 27: What is the unit of power related to work?

Answer: Watt (W). One watt equals one joule per second.

Flashcard 28: If displacement is zero, what is work done regardless of force?

Answer: Zero work. Work formula requires displacement to be nonzero.

Flashcard 29: What is the effect of angle on work done?

Answer: Work depends on $\text{cos}(\theta)$; maximum at $0^\text{o}$, zero at $90^\text{o}$. The cosine function determines the effective force component.

Flashcard 30: Calculate work: $F = 15 \text{ N}$, $d = 4 \text{ m}$, $\theta = 180^\text{o}$.

Answer: -60 J. Using $\cos(180°) = -1$, so $W = 15 × 4 × (-1) = -60$ J.