AP Physics 1 Quiz: Work

What this quiz covers

This quiz focuses on Work, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Zero, because the track is horizontal
2. Negative, because the cart moves left
3. Positive (correct answer)
4. Cannot be determined without the cart’s speed

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by an applied force. Work is defined as the dot product of force and displacement, W = F · d = F d cosθ, where θ is the angle between the force and displacement vectors. Here, both the applied force and displacement are to the left, so θ = 0° and cosθ = 1, yielding positive work. Qualitatively, when a force acts in the same direction as displacement, it does positive work by adding energy to the system. A common distractor is choice B, which wrongly attributes negative work to the leftward motion, but direction labels like 'left' do not inherently make work negative. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 2

1. Positive, because gravity has constant magnitude
2. Zero, because the ball is moving upward
3. Negative, because gravity is opposite the displacement (correct answer)
4. Positive, because work depends only on distance traveled

Explanation: This question tests understanding of work done by gravity on rising objects. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement. The ball moves upward while gravity acts downward, making θ = 180°. Since cos(180°) = -1, the work is negative: W = mg·d·(-1) < 0. Choice D incorrectly suggests work depends only on distance magnitude, ignoring the crucial directional relationship. Gravity does negative work on any object moving upward against it.

Question 3

1. Positive, because a force is applied while the box moves
2. Zero, because the force is perpendicular to the displacement (correct answer)
3. Negative, because the force is not in the direction of motion
4. $30\,\text{J}$, because $W=Fd$

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on work done by a force perpendicular to displacement. Work is the dot product of force and displacement, W = F · d = F d cosθ, where θ determines the component of force along the displacement. If the force is perpendicular to the displacement, θ = 90°, cosθ = 0, so work is zero. Here, the upward force is perpendicular to the rightward displacement, resulting in zero work. Choice A is a distractor because it ignores the directional requirement, assuming any force during motion does positive work. A useful strategy is to visualize the force and displacement vectors to quickly identify if they are parallel, antiparallel, or perpendicular.

Question 4

1. Negative, because leftward forces do negative work
2. Zero, because the displacement is only $2.0\,\text{m}$
3. Positive, because force and displacement are in the same direction (correct answer)
4. Zero, because the force is constant

Explanation: This question tests understanding of work when force and displacement align. Work is W = F·d·cos(θ), where θ is the angle between force and displacement vectors. Both the force and cart displacement are to the left, making θ = 0°. Since cos(0°) = 1, the work is positive: W = (7.0 N)(2.0 m)(1) = +14 J. Choice A incorrectly assumes leftward forces always do negative work, confusing force direction with the force-displacement relationship. When force and displacement point the same way, work is positive regardless of compass direction.

Question 5

1. Positive
2. Negative (correct answer)
3. Zero, because the box moves at constant height
4. Cannot be determined without the box’s mass

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by friction. Work is defined as the dot product of force and displacement, W = F · d = F d cosθ, where θ is the angle between the force and displacement vectors. In this case, the friction force acts to the left while the displacement is to the right, making θ = 180° and cosθ = -1, resulting in negative work. Qualitatively, when a force opposes the displacement, it does negative work by removing energy from the system. A common distractor is choice C, which incorrectly assumes zero work due to constant height, ignoring that friction is horizontal and opposite to motion. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 6

1. Positive, because the puck’s displacement is to the right
2. Negative, because the force is opposite the displacement (correct answer)
3. Zero, because friction is negligible
4. Positive, because any nonzero force does positive work

Explanation: This question tests work concepts in AP Physics 1, emphasizing negative work from opposing forces. Work is the dot product W = F · d = F d cosθ, resulting in negative values for θ = 180° where cosθ = -1. The leftward force opposes the rightward displacement of the puck, yielding negative work. This slows the puck despite low friction. Choice A incorrectly assumes positive work based on displacement direction alone, ignoring the force's opposition. A practical strategy is to use the formula's cosine term to systematically determine work's sign in any orientation.

Question 7

1. Positive, because the hand exerts a force while moving
2. Negative, because gravity opposes the hand’s force
3. Zero, because the force is perpendicular to the displacement (correct answer)
4. Nonzero, because constant speed implies constant work

Explanation: This question tests work in AP Physics 1, specifically when a supporting force is perpendicular to displacement. Work is defined by the dot product W = F · d = F d cosθ, where perpendicular vectors give θ = 90° and cosθ = 0, hence zero work. The upward force here does not contribute along the horizontal displacement direction. Thus, the work by the hand's upward force is zero. Choice A is a common distractor, assuming motion with force implies positive work without considering direction. A transferable approach is to decompose forces into components parallel and perpendicular to displacement before calculating work.

Question 8

1. Zero, because the cart is moving
2. Positive, because the force is $15\,\text{N}$
3. Negative (correct answer)
4. Zero, because the force opposes motion

Explanation: This question tests understanding of work as the dot product of force and displacement. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The cart moves 6.0 m right while the force acts 15 N left, making these vectors opposite (θ = 180°). Since cos(180°) = -1, the work is negative: W = (15 N)(6.0 m)(-1) = -90 J. Choice B incorrectly assumes positive work just because the force magnitude is positive, ignoring the crucial directional relationship. When force opposes displacement, work is always negative.

Question 9

1. Positive, because friction is a force
2. Negative, because friction opposes the displacement (correct answer)
3. Zero, because the sled moves horizontally
4. Positive, because the sled moves $8.0\,\text{m}$

Explanation: This question tests understanding of work done by friction. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement. The sled moves forward while friction acts backward, making θ = 180°. Since cos(180°) = -1, the work is negative: W = (5.0 N)(8.0 m)(-1) = -40 J. Choice A incorrectly assumes all forces do positive work, ignoring the opposition between friction and motion. Friction opposing motion always does negative work on the moving object.

Question 10

1. Cannot be determined without the time interval
2. Zero, because work depends only on displacement
3. Negative, because the force is constant
4. Positive (correct answer)

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by a constant horizontal force. Work is defined as the dot product of force and displacement, W = F · d = F d cosθ, where θ is the angle between the force and displacement vectors. Both force and displacement are to the right, so θ = 0° and cosθ = 1, giving positive work. Qualitatively, a force in the direction of displacement adds energy, performing positive work. A common distractor is choice C, which wrongly suggests negative work due to the force being constant, but constancy does not affect the sign. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 11

1. Negative (correct answer)
2. Positive, because tension is a pulling force
3. Zero, because the force is constant
4. Cannot be determined without the box’s acceleration

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by tension. Work is defined as the dot product of force and displacement, W = F · d = F d cosθ, where θ is the angle between the force and displacement vectors. Tension acts to the right while displacement is to the left, making θ = 180° and cosθ = -1, so work is negative. Qualitatively, a force opposite to displacement does negative work, removing energy. A common distractor is choice B, which wrongly assumes positive work because tension is a pulling force, ignoring its opposition to motion. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 12

1. Positive, because the force is $2.0\,\text{N}$
2. Negative (correct answer)
3. Zero, because ice implies no work is done
4. Cannot be determined without the puck’s mass

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by a horizontal force on a puck. Work is defined as the dot product of force and displacement, W = F · d = F d cosθ, where θ is the angle between the force and displacement vectors. The force is to the right while displacement is to the left, so θ = 180° and cosθ = -1, resulting in negative work. Qualitatively, when force opposes displacement, it does negative work by reducing kinetic energy. A common distractor is choice A, which incorrectly claims positive work based on force magnitude alone, without considering direction. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 13

1. Positive, because the force has nonzero magnitude
2. Negative, because the force is opposite the displacement (correct answer)
3. Zero, because the cart moves at constant height
4. Positive, because work equals force

Explanation: This question tests understanding of work as the dot product of force and displacement. Work is calculated as W = F·d·cos(θ), where θ is the angle between the force and displacement vectors. Here, the cart moves 5.0 m to the right while the force acts to the left, making θ = 180°. Since cos(180°) = -1, the work is negative: W = (3.0 N)(5.0 m)(-1) = -15 J. Choice A incorrectly assumes any nonzero force does positive work, ignoring direction. When force opposes displacement, work is always negative.

Question 14

1. Negative, because the force is opposite the displacement (correct answer)
2. Positive, because the book moves to the right
3. Zero, because the force is constant
4. Positive, because the force magnitude is $10\,\text{N}$

Explanation: This question probes the concept of work in AP Physics 1, emphasizing negative work when force opposes displacement. Work is the dot product of force and displacement, W = F · d = F d cosθ, yielding negative values when θ = 180° and cosθ = -1. This occurs when the force acts against the direction of motion, extracting energy. Here, the hand's force to the left opposes the book's rightward displacement, resulting in negative work. Choice B incorrectly focuses on the displacement direction alone, ignoring the relative orientation to the force. To avoid errors, consistently evaluate the cosine of the angle between force and displacement for any work calculation.

Question 15

1. Positive, because the force is $9.0\,\text{N}$
2. Zero, because the motion is one-dimensional
3. Negative, because the force is opposite the displacement (correct answer)
4. Zero, because the force is constant

Explanation: This question tests understanding of work with opposing force and displacement. Work is calculated as W = F·d·cos(θ), where θ is the angle between the vectors. The cart moves left while the force acts right, making θ = 180°. Since cos(180°) = -1, the work is W = (9.0 N)(3.5 m)(-1) = -31.5 J, which is negative. Choice A incorrectly focuses on force magnitude rather than direction, missing that work's sign depends on the relative directions of force and displacement. Remember that work is negative when force opposes displacement, regardless of which direction is labeled positive.

Question 16

1. Positive, because the normal force is $30\,\text{N}$
2. Negative, because the force is upward
3. Zero (correct answer)
4. Positive, because the puck moved $1.2\,\text{m}$

Explanation: This question tests understanding of work as the dot product of force and displacement. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The puck moves 1.2 m left (horizontal) while the normal force acts 30 N upward (vertical), making these vectors perpendicular (θ = 90°). Since cos(90°) = 0, the work is zero: W = (30 N)(1.2 m)(0) = 0 J. Choice B incorrectly assumes upward forces do negative work on horizontal motion, but perpendicular forces do no work. When force is perpendicular to displacement, work is always zero.

Question 17

1. Positive, because friction is a force
2. Negative (correct answer)
3. Zero, because friction is constant
4. Zero, because the sled moves downhill

Explanation: This question tests understanding of work as the dot product of force and displacement. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The sled slides 8.0 m down the hill while friction acts 10 N up the hill, making these vectors opposite (θ = 180°). Since cos(180°) = -1, the work is negative: W = (10 N)(8.0 m)(-1) = -80 J. Choice C incorrectly suggests constant friction means zero work, but work depends on the force-displacement angle, not force constancy. Friction opposing motion always does negative work.

Question 18

1. Positive, because the force is $20\,\text{N}$
2. Negative, because the force is not in the direction of motion
3. Zero (correct answer)
4. Positive, because the crate moved $3.0\,\text{m}$

Explanation: This question tests understanding of work as the dot product of force and displacement. Work is W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The crate moves 3.0 m left (horizontal) while the force acts 20 N upward (vertical), making these vectors perpendicular (θ = 90°). Since cos(90°) = 0, the work is zero: W = (20 N)(3.0 m)(0) = 0 J. Choice B incorrectly assumes any force not aligned with motion produces negative work, but perpendicular forces do zero work. When force is perpendicular to displacement, no work is done regardless of the magnitudes.

Question 19

1. Negative, because spring forces always do negative work
2. Positive, because the force is in the direction of displacement (correct answer)
3. Zero, because the force is constant
4. Zero, because only net force can do work

Explanation: This question examines work in AP Physics 1, focusing on positive work from aligned force and displacement. Work is the dot product W = F · d = F d cosθ, positive when θ = 0° as cosθ = 1. This means the force adds energy in the direction of motion. For the toy car, the spring force to the right matches the rightward displacement, yielding positive work. Choice A wrongly claims spring forces always do negative work, confusing restoring forces with direction-specific calculations. Remember to assess each situation individually by comparing force and displacement directions for accurate sign determination.

Question 20

1. Zero, because the table prevents vertical motion
2. Negative, because the force is applied
3. Positive (correct answer)
4. Zero, because work depends only on force magnitude

Explanation: This question tests understanding of work as the dot product of force and displacement. Work is calculated as W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The book moves 2.5 m right and the applied force acts 9.0 N right, making these vectors parallel (θ = 0°). Since cos(0°) = 1, the work is positive: W = (9.0 N)(2.5 m)(1) = +22.5 J. Choice A incorrectly focuses on the table's role in preventing vertical motion, which is irrelevant to the horizontal work calculation. When force and displacement align, work is always positive.