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.

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.

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.

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.

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.

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.

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.

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.

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.

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.