AP Physics 1 Quiz: Work
20 questions · exam conditions
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WorkQuestion 1 of 20

A cart moves 3.0m3.0\,\text{m} to the left on a horizontal track. A constant applied force of 4.0N4.0\,\text{N} acts to the left, in the same direction as the displacement. What is the sign of the work done by the applied force on the cart?

Zero, because the track is horizontal
Negative, because the cart moves left
Positive
Cannot be determined without the cart's speed
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AP Physics 1 Quiz

AP Physics 1 Quiz: Work

Practice Work in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

A cart moves 3.0m3.0\,\text{m} to the left on a horizontal track. A constant applied force of 4.0N4.0\,\text{N} acts to the left, in the same direction as the displacement. What is the sign of the work done by the applied force on the cart?

  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

A ball moves straight upward 3.0m3.0\,\text{m} after release while the constant gravitational force acts downward. What is the sign of the work done by gravity on the ball during this displacement?

  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

A student pushes a box 2.0m2.0\,\text{m} to the right along a level floor. During this displacement, a constant horizontal friction force of 5.0N5.0\,\text{N} acts on the box to the left, opposite the displacement. The student's push is not considered. What is the sign of the work done by friction on the box?

  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 4

A puck slides 7.0m7.0\,\text{m} to the right on nearly frictionless ice. A constant 4.0N4.0\,\text{N} force acts to the left on the puck, opposite the displacement. What is the sign of the work done by this force?

  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 5

A suitcase is carried 9.0m9.0\,\text{m} horizontally to the right at constant height. The constant upward force from the person's hand supports the suitcase, perpendicular to the displacement. What is the work done by the hand's upward force?

  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 6

A cart moves 5.0m5.0\,\text{m} to the right. A constant horizontal force of 12N12\,\text{N} acts to the right, in the same direction as the displacement. What is the sign of the work done by this force?

  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 7

A box is pulled 7.0m7.0\,\text{m} to the left. A constant horizontal tension force of 9.0N9.0\,\text{N} acts to the right, opposite the displacement. What is the sign of the work done by the tension force?

  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 8

A 2.0kg2.0\,\text{kg} cart moves 5.0m5.0\,\text{m} to the right on a level track while a constant 3.0N3.0\,\text{N} force acts to the left. What is the sign of the work done by this force on the cart?

  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 9

A book is pushed 1.8m1.8\,\text{m} to the right across a table. A constant 10N10\,\text{N} force from a hand acts to the left, opposite the displacement. What is the sign of the work done by the hand on the book?

  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 10N10\,\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 10

A sled slides 8.0m8.0\,\text{m} down a straight hill. A constant friction force of 10N10\,\text{N} acts up the hill, opposite the displacement. What is the sign of the work done by friction?

  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 11

A toy car rolls 3.0m3.0\,\text{m} to the right. A constant spring force of 5.0N5.0\,\text{N} acts to the right during the motion, in the same direction as displacement. What is the sign of the work done by the spring force?

  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 12

A block slides 4.0m4.0\,\text{m} to the right on a rough surface. A constant kinetic friction force acts 8.0N8.0\,\text{N} to the left, opposite the displacement. What is the sign of the work done by friction on the block?

  1. Negative (correct answer)
  2. Positive, because friction has a nonzero magnitude
  3. Zero, because friction is not a conservative force
  4. Cannot be determined without the coefficient of friction

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by kinetic 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. The friction force is to the left and displacement to the right, making θ = 180° and cosθ = -1, so work is negative. Qualitatively, friction opposes motion, doing negative work by dissipating energy as heat. A common distractor is choice B, which incorrectly assumes positive work from nonzero friction magnitude, disregarding 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

A book slides 1.8m1.8\,\text{m} to the right on a table while a constant normal force acts upward. What is the work done by the normal force on the book?

  1. Positive, because the normal force is large
  2. Negative, because the normal force opposes motion
  3. Zero, because the normal force is perpendicular to the displacement (correct answer)
  4. Nonzero, because any force does work during motion

Explanation: This question tests understanding of work when force is perpendicular to displacement. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The normal force acts upward while the book slides horizontally, making θ = 90°. Since cos(90°) = 0, the work is W = F·d·(0) = 0 J regardless of the magnitudes of force or displacement. Choice D incorrectly assumes any force during motion does work, but perpendicular forces do zero work. To quickly identify zero work situations, check if the force is perpendicular to the displacement direction.

Question 14

A sled moves 8.0m8.0\,\text{m} forward while kinetic friction of constant magnitude 5.0N5.0\,\text{N} acts backward. What is the sign of the work done by friction on the sled?

  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 8.0m8.0\,\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 15

A cart moves 3.5m3.5\,\text{m} to the left while a constant 9.0N9.0\,\text{N} force acts to the right. What is the sign of the work done by the force?

  1. Positive, because the force is 9.0N9.0\,\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

A book is pushed 2.5m2.5\,\text{m} to the right across a table. A constant applied force of 9.0N9.0\,\text{N} acts to the right during the motion. What is the sign of the work done by the applied force?

  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.

Question 17

A sled is pulled 6.0m6.0\,\text{m} to the right across snow. The only force considered is a constant 10N10\,\text{N} force exerted by the rope to the left, opposite the displacement. What is the sign of the work done by the rope on the sled?

  1. Positive, because the force magnitude is 10N10\,\text{N}
  2. Negative (correct answer)
  3. Zero, because the force is horizontal
  4. Zero, because the sled moves at constant speed

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by a rope 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. The rope force acts to the left while the displacement is to the right, so θ = 180° and cosθ = -1, resulting in negative work. Qualitatively, a force opposing the displacement performs negative work, extracting energy from the system. A common distractor is choice A, which mistakenly claims positive work based solely on force magnitude, ignoring the directional relationship. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 18

A robot drags a bin 8.0m8.0\,\text{m} to the right across a floor. A constant resistive force of 6.0N6.0\,\text{N} acts to the left, opposite the displacement, and no other forces are considered. What is the sign of the work done by the resistive force?

  1. Cannot be determined without the bin's speed
  2. Zero, because resistive forces are not real forces
  3. Negative (correct answer)
  4. Positive, because the bin moves 8.0m8.0\,\text{m}

Explanation: This question assesses the understanding of work in AP Physics 1, focusing on determining the sign of work done by a resistive 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. The resistive force is to the left while displacement is to the right, making θ = 180° and cosθ = -1, so work is negative. Qualitatively, resistive forces oppose motion, doing negative work by dissipating energy. A common distractor is choice D, which mistakenly suggests positive work based on displacement distance, without regard to force direction. To determine the sign of work in similar problems, always check if the force is parallel, antiparallel, or perpendicular to the displacement.

Question 19

A student pulls a sled 4.0m4.0\,\text{m} to the right on level snow. A constant 30N30\,\text{N} friction force acts to the left, opposite the displacement. What is the sign of the work done on the sled by friction?

  1. Positive, because the friction force has magnitude 30N30\,\text{N}
  2. Zero, because friction is not an applied force
  3. Negative, because the friction force is opposite the displacement (correct answer)
  4. Positive, because the sled moves to the right

Explanation: This question tests the concept of work in AP Physics 1, specifically the sign of work done by friction on a moving object. Work is defined as the dot product of force and displacement vectors, W = F · d = F d cosθ, where θ is the angle between them. When the force is opposite to the displacement, θ = 180°, cosθ = -1, resulting in negative work. In this case, the friction force acts left while the sled moves right, so the work by friction is negative. Choice A is incorrect because the sign of work depends on the direction relative to displacement, not just the magnitude. To determine the sign of work in any scenario, always check the angle between the force and displacement vectors.

Question 20

A student lifts a 2.0kg2.0\,\text{kg} object straight up 0.80m0.80\,\text{m}. The constant gravitational force on the object acts downward, opposite the displacement. What is the sign of the work done by gravity?

  1. Zero, because gravity is constant
  2. Positive, because the object moves upward
  3. Negative, because gravity is opposite the displacement (correct answer)
  4. Positive, because the object has weight

Explanation: This question evaluates understanding of work in AP Physics 1, focusing on gravitational work during lifting. Work is computed as W = F · d = F d cosθ, with negative sign when force is antiparallel to displacement (θ = 180°). Gravity pulls downward while the object moves upward, so work by gravity is negative. This represents energy transfer out of the system. Choice B mistakenly ties positive work to upward motion without considering force direction. To generalize, always identify if a force aids or opposes motion to predict work's sign effectively.