Work - AP Physics 1

1. A puck slides $7.0\,\text{m}$ to the right on nearly frictionless ice. A constant $4.0\,\text{N}$ force acts to the left on the puck, opposite the displacement. What is the sign of the work done by this force?
   Answer: Negative, because the force is opposite the displacement 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.
2. A book slides $1.8\,\text{m}$ 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?
   Answer: Zero, because the normal force is perpendicular to the displacement 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.
3. A student pushes a box $2.0\,\text{m}$ to the right along a level floor. During this displacement, a constant horizontal friction force of $5.0\,\text{N}$ 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?
   Answer: Negative 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.
4. A sled is pulled $6.0\,\text{m}$ to the right across snow. The only force considered is a constant $10\,\text{N}$ 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?
   Answer: Negative 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.
5. A block slides $0.50\,\text{m}$ to the right while a constant $4.0\,\text{N}$ force acts to the left. What is the sign of the work done by this force?
   Answer: Negative, because the force is opposite the displacement This question tests understanding of work when force opposes displacement. Work is W = F·d·cos(θ), where θ is the angle between force and displacement. The block slides right while the force acts left, making θ = 180°. Since cos(180°) = -1, the work is negative: W = (4.0 N)(0.50 m)(-1) = -2.0 J. Choice B incorrectly assumes the force magnitude makes work positive, ignoring direction. When force opposes displacement, work is always negative regardless of the specific values.
6. A cart moves $6.0\,\text{m}$ to the right. A constant $15\,\text{N}$ force acts to the left on the cart during this displacement. What is the sign of the work done by this force?
   Answer: Negative 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.
7. A cart moves $2.0\,\text{m}$ to the left while a constant $7.0\,\text{N}$ force acts to the left. What is the sign of the work done by this force on the cart?
   Answer: Positive, because force and displacement are in the same direction 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.
8. A student pulls a sled $4.0\,\text{m}$ to the right on level snow. A constant $30\,\text{N}$ friction force acts to the left, opposite the displacement. What is the sign of the work done on the sled by friction?
   Answer: Negative, because the friction force is opposite the displacement 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.
9. A robot drags a crate $3.0\,\text{m}$ west while a constant $9.0\,\text{N}$ force acts east. What is the sign of the work done by the $9.0\,\text{N}$ force?
   Answer: Negative, because the force is opposite the displacement This question tests understanding of work when force opposes displacement. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement vectors. The crate moves west while the 9.0 N force acts east (opposite directions), making θ = 180°. Since cos(180°) = -1, the work is negative: W = (9.0 N)(3.0 m)(-1) = -27 J. Choice A incorrectly focuses only on force magnitude, ignoring the opposition to motion. When analyzing work, always check if force and displacement are in the same or opposite directions.
10. A cart moves $3.5\,\text{m}$ to the left while a constant $9.0\,\text{N}$ force acts to the right. What is the sign of the work done by the force?
    Answer: Negative, because the force is opposite the displacement 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.
11. A suitcase is pulled $10\,\text{m}$ forward while a constant $20\,\text{N}$ handle force acts at $90^\circ$ to the motion (sideways). What is the work done by the handle force?
    Answer: Zero, because the force is perpendicular to the displacement 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. The suitcase moves forward while the handle force acts sideways at 90°, so θ = 90°. Since cos(90°) = 0, the work is zero: W = (20 N)(10 m)(0) = 0 J. Choice A incorrectly focuses on force magnitude alone, missing the perpendicular relationship. When force is perpendicular to displacement, no work is done regardless of magnitudes.
12. A robot pulls a wagon $5.0\,\text{m}$ to the left. A constant tension force of $18\,\text{N}$ acts to the left on the wagon. What is the sign of the work done by the tension force?
    Answer: Positive 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 wagon moves 5.0 m left and the tension force acts 18 N left, making these vectors parallel (θ = 0°). Since cos(0°) = 1, the work is positive: W = (18 N)(5.0 m)(1) = +90 J. Choice A incorrectly assumes leftward forces do negative work, but direction signs depend on the force-displacement relationship, not absolute directions. When force and displacement point the same way, work is positive.
13. A sled moves $10\,\text{m}$ to the right while a constant wind force of $4.0\,\text{N}$ acts to the right. What is the sign of the work done by the wind on the sled?
    Answer: Positive, because the force is in the direction of displacement This question tests understanding of work by assisting forces. Work equals W = F·d·cos(θ), where θ is the angle between force and displacement. Both the wind force and sled displacement are to the right, making θ = 0°. Since cos(0°) = 1, the work is W = (4.0 N)(10 m)(1) = 40 J, which is positive. Choice A incorrectly assumes wind always does negative work as a resistive force, but here the wind assists motion in the same direction. When force aids motion in the displacement direction, work is positive regardless of the force's nature.
14. A book is pushed $1.8\,\text{m}$ to the right across a table. A constant $10\,\text{N}$ 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?
    Answer: Negative, because the force is opposite the displacement 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.
15. A box slides $2.5\,\text{m}$ to the right on a horizontal floor. A constant applied force of $12\,\text{N}$ acts upward, perpendicular to the displacement. What is the work done by this applied force?
    Answer: Zero, because the force is perpendicular to the displacement 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.
16. A $2.0\,\text{kg}$ cart moves $5.0\,\text{m}$ to the right on a level track while a constant $3.0\,\text{N}$ force acts to the left. What is the sign of the work done by this force on the cart?
    Answer: Negative, because the force is opposite the displacement 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.
17. A cart moves $6.0\,\text{m}$ to the right while a constant friction force of $3.0\,\text{N}$ acts to the left. What is the sign of the work done by friction on the cart?
    Answer: Negative, because friction is opposite the displacement This question tests understanding of work done by friction forces. Work is calculated as W = F·d·cos(θ), where θ is the angle between force and displacement. The cart moves right while friction acts left, making θ = 180°. Since cos(180°) = -1, the work is W = (3.0 N)(6.0 m)(-1) = -18 J, which is negative. Choice D incorrectly assumes constant speed means zero work, but work depends on force and displacement, not velocity. Friction typically does negative work on moving objects because it opposes the direction of motion.
18. A box is pulled $4.0\,\text{m}$ east while a constant $10\,\text{N}$ force acts north. What is the work done by this force on the box?
    Answer: Zero, because the force is perpendicular to the displacement 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. The box moves east while the force acts north, making θ = 90°. Since cos(90°) = 0, the work is zero: W = (10 N)(4.0 m)(0) = 0 J. Choice B incorrectly assumes the force magnitude alone determines work, ignoring the crucial directional relationship. When force is perpendicular to displacement, no work is done regardless of force magnitude.
19. A puck slides $1.5\,\text{m}$ to the left on ice. A constant horizontal force of $2.0\,\text{N}$ acts to the right, opposite the displacement. What is the sign of the work done by that force?
    Answer: Negative 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.
20. A book is pushed $2.5\,\text{m}$ to the right across a table. A constant applied force of $9.0\,\text{N}$ acts to the right during the motion. What is the sign of the work done by the applied force?
    Answer: Positive 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.

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