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 dot product of vectors in the work formula?

Answer: The dot product is $F \times d \times \text{cos}(\theta)$. The mathematical operation for work between two vectors.

Flashcard 2: Define the term 'conservative force'.

Answer: A force with work independent of path, e.g., gravity. Work depends only on initial and final positions.

Flashcard 3: What is the work done by gravity on an object moving horizontally?

Answer: Zero, as displacement is perpendicular to gravity. Gravity acts vertically while motion is horizontal.

Flashcard 4: Define work done by a non-conservative force.

Answer: Work depends on the path taken. Examples include friction and air resistance forces.

Flashcard 5: 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 6: 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 7: What is the work done by a conservative force over a closed path?

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

Flashcard 8: 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 9: What is the work done by friction?

Answer: Negative, as friction opposes motion. Friction force acts opposite to the direction of motion.

Flashcard 10: 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 11: Calculate power: $W = 120 \text{ J}$, time $t = 4 \text{ s}$.

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

Flashcard 12: Define negative work.

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

Flashcard 13: 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 14: State the SI unit of work.

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

Flashcard 15: 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 16: Define work in terms of force and displacement.

Answer: Work is the product of force and displacement in the direction of force. Only the component of force parallel to displacement contributes to work.

Flashcard 17: What is the unit of power related to work?

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

Flashcard 18: If displacement is zero, what is work done regardless of force?

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

Flashcard 19: 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 20: 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 21: State the work-energy theorem.

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

Flashcard 22: 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 23: Define work done by a non-conservative force.

Answer: Work depends on the path taken. Examples include friction and air resistance forces.

Flashcard 24: 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 25: 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 26: 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.

Flashcard 27: 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 28: What is the dot product of vectors in the work formula?

Answer: The dot product is $F \times d \times \text{cos}(\theta)$. The mathematical operation for work between two vectors.

Flashcard 29: 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 30: 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 31: 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 32: 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 33: 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 34: Define work in terms of force and displacement.

Answer: Work is the product of force and displacement in the direction of force. Only the component of force parallel to displacement contributes to work.

Flashcard 35: Calculate work: $F = 10 \text{ N}$, $d = 10 \text{ m}$, $\theta = 45^\text{o}$.

Answer: 70.7 J. Using $\cos(45°) = \frac{\sqrt{2}}{2} ≈ 0.707$, so $W = 10 × 10 × 0.707 = 70.7$ J.

Flashcard 36: 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 37: 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 38: For $W = 75 \text{ J}$, $F = 15 \text{ N}$, find $d$ when $\theta = 0^\text{o}$.

Answer: $d = 5 \text{ m}$. Solving $d = W/F = 75/15 = 5$ m.

Flashcard 39: What is an example of a non-conservative force?

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

Flashcard 40: 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 41: 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 42: Calculate power: $W = 120 \text{ J}$, time $t = 4 \text{ s}$.

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

Flashcard 43: Identify the scalar quantity: work or force?

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

Flashcard 44: 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 45: 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 46: What is the work done by gravity on an object moving horizontally?

Answer: Zero, as displacement is perpendicular to gravity. Gravity acts vertically while motion is horizontal.

Flashcard 47: What is the work done when no displacement occurs?

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

Flashcard 48: 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 49: 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 50: 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.