This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i, where work is calculated as W = Fd for a constant force parallel to displacement (positive if in direction of motion, negative if opposing). In this scenario, an applied force of 40 N acts to the right and friction of 25 N acts to the left over displacement d = 6.0 m to the right. The work done by the applied force is W_applied = (40 N)(6.0 m) = +240 J (positive because force and displacement are in same direction), and the work done by friction is W_friction = -(25 N)(6.0 m) = -150 J (negative because friction opposes motion). The net work is W_net = W_applied + W_friction = 240 J + (-150 J) = +90 J. Choice B (+90 J) is correct because it properly accounts for the positive work by the applied force and the negative work by friction. Choice C (+240 J) reports only the work done by the applied force, ignoring the negative work done by friction—the question asks for net work, which requires summing all forces' contributions. When calculating net work: (1) identify all forces and their directions relative to displacement, (2) calculate work by each force: positive if in direction of motion, negative if opposing, (3) sum all work contributions algebraically to get W_net.

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement, and kinetic energy is KE = ½mv². To stop the cart (v_f = 0) from initial velocity v_i = 12 m/s, we calculate the change in kinetic energy: ΔKE = 0 - ½m(v_i²) = -½(2.0 kg)(12 m/s)² = -½(2.0)(144) = -144 J (negative because kinetic energy decreases). The only force doing work is friction, which opposes motion, so W_net = W_friction = -f × d = -(4.0 N) × d. Using W_net = ΔKE: -(4.0 N) × d = -144 J, solving for d: d = 144 J / 4.0 N = 36 m. Choice B (36 m) is correct because it properly applies the work-energy theorem with the correct kinetic energy calculation including the ½ factor. Choice C (72 m) is incorrect—it appears to result from forgetting the ½ factor in kinetic energy, calculating ΔKE = -mv² = -2.0(144) = -288 J, which doubles the stopping distance. When finding stopping distance: (1) calculate initial kinetic energy KE_i = ½mv_i² (don't forget the ½), (2) recognize that all this energy must be removed by friction work, so W_friction = -KE_i, (3) use W_friction = -fd to solve for distance d.

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). For applying work-energy to find velocity: the net work done on the object is W_net = -350 J; using W_net = ½m(v_f² - v_i²) with m = 7.0 kg and v_i = 10 m/s: -350 J = ½(7)(v_f² - 100), solving for v_f: v_f² = 100 + (-350*2)/7 = 100 - 700/7 = 100 - 100 = 0, so v_f = √0 = 0 m/s. Choice D is correct because it accurately applies the work-energy theorem to find that the negative net work brings the cart to a stop. Choice C treats the work as positive, leading to v_f = √(100 + 700/7) = √200 ≈ 14.1 m/s but adjusted, or miscalculates the sign. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the ½ in KE = ½mv² (makes energy twice too large), (b) using v instead of v² (dramatically underestimates KE), (c) treating friction or braking forces as doing positive work when they oppose motion (always W_friction = -f·d), and (d) confusing individual work with net work (must sum all forces' work to get W_net).

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). For calculating kinetic energy change: the object has mass m = 4.0 kg, initial velocity v_i = 3.0 m/s, and final velocity v_f = 9.0 m/s; the change in kinetic energy is ΔKE = ½m(v_f² - v_i²) = ½(4)((9)² - (3)²) = 2*(81 - 9) = 2*72 = 144 J; positive ΔKE means the object gained kinetic energy (sped up). Choice A is correct because it correctly calculates ΔKE = ½mv_f² - ½mv_i² with velocity squared and the ½ factor. Choice B forgets the ½ factor in the kinetic energy formula, calculating ΔKE = m(v_f² - v_i²)/2 but wait, actually B is half of 144, perhaps calculating only final or initial KE. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the ½ in KE = ½mv² (makes energy twice too large), (b) using v instead of v² (dramatically underestimates KE), (c) treating friction or braking forces as doing positive work when they oppose motion (always W_friction = -f·d), and (d) confusing individual work with net work (must sum all forces' work to get W_net).

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). In this scenario, a constant applied force of magnitude F = 70 N acts over displacement d = 8.0 m in the same direction while friction of 25 N opposes motion; the net work is W_net = W_applied + W_friction = (70)(8) + (-25)(8) = 560 - 200 = 360 J. Choice B is correct because it accurately calculates net work by summing the positive work by the applied force and negative work by friction. Choice A incorrectly adds the force magnitudes before multiplying by displacement, treating friction as positive work instead of negative. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the negative sign for opposing forces like friction, (b) confusing individual work with net work, and (c) ignoring the direction of force relative to displacement.

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). In this scenario, a constant applied force of magnitude F = 100 N acts over displacement d = 6.0 m at angle θ = 30° above horizontal; the work done is W = Fd cos(θ) = (100)(6) cos(30°) = 600 * (√3/2) ≈ 600 * 0.866 = 519.6 ≈ 520 J. Choice B is correct because it uses the work formula correctly with cos(θ) to account for the horizontal component of the force. Choice A incorrectly uses W = Fd without cos(θ), assuming the full force is parallel to displacement when it's at an angle. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the ½ in KE = ½mv² (makes energy twice too large), (b) using v instead of v² (dramatically underestimates KE), (c) treating friction or braking forces as doing positive work when they oppose motion (always W_friction = -f·d), and (d) confusing individual work with net work (must sum all forces' work to get W_net).

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). For determining force from work-energy: to stop an object (v_f = 0) from initial velocity v_i = 22 m/s over distance d = 55 m, we use W_net = ΔKE; the change in kinetic energy is ΔKE = 0 - ½m(v_i²) = -½(1200)(22)² = -600*484 = -290400 J (negative because kinetic energy decreases); since W_net = -F × d and W_net = ΔKE, we have -F × 55 = -290400 J, giving F = 290400/55 ≈ 5280 N ≈ 5.3×10³ N in magnitude (negative sign indicates force opposes motion). Choice A is correct because it correctly accounts for the sign of work by the braking force (negative) and applies the work-energy theorem to find the force magnitude. Choice B treats the kinetic energy as mv² instead of ½mv², doubling the energy and thus the force. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the ½ in KE = ½mv² (makes energy twice too large), (b) using v instead of v² (dramatically underestimates KE), (c) treating friction or braking forces as doing positive work when they oppose motion (always W_friction = -f·d), and (d) confusing individual work with net work (must sum all forces' work to get W_net).

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). In this scenario, kinetic friction of magnitude F = 15 N acts over displacement d = 12 m opposing motion; the work done is W = -Fd = -15*12 = -180 J. Choice A is correct because it correctly accounts for the sign of work: negative for friction opposing motion. Choice B treats the friction work as positive instead of negative—friction opposes motion so its work is W_friction = -f·d (negative), removing kinetic energy from the object. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Key relationships: if net work is positive, the object speeds up (gains kinetic energy); if net work is negative, the object slows down (loses kinetic energy); if net work is zero, speed stays constant—and the work-energy theorem provides a powerful alternative to F = ma when dealing with problems involving displacement and velocity changes.

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). In this scenario, a constant applied force of magnitude F = 40 N acts over displacement d = 4.0 m in the same direction while friction of 10 N opposes motion; the net work is W_net = W_applied + W_friction = (40)(4) + (-10)(4) = 160 - 40 = 120 J. Choice B is correct because it accurately calculates net work by summing the positive work by the applied force and negative work by friction. Choice A forgets the negative sign for friction, adding instead of subtracting the works. When solving work-energy problems: (1) identify all forces and calculate work by each: W = Fd if parallel (positive), W = -Fd if opposing (negative), W = Fd cos(θ) if at angle, (2) find net work by summing: W_net = W₁ + W₂ + W₃ + ..., (3) calculate initial and final kinetic energies: KE = ½mv² (don't forget the ½ and v²), then (4) apply work-energy theorem: W_net = ΔKE = KE_f - KE_i to solve for the unknown. Common errors to avoid: (a) forgetting the ½ in KE = ½mv² (makes energy twice too large), (b) using v instead of v² (dramatically underestimates KE), (c) treating friction or braking forces as doing positive work when they oppose motion (always W_friction = -f·d), and (d) confusing individual work with net work (must sum all forces' work to get W_net).

This question tests understanding of the work-energy theorem, which relates the net work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_f - KE_i = ½m(v_f² - v_i²), where work is calculated as W = Fd for a constant force parallel to displacement (or W = Fd cos(θ) if force at angle θ), and kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). For calculating kinetic energy change: The object has mass m = 5.0 kg, initial velocity v_i = 4.0 m/s, and final velocity v_f = 10.0 m/s. The change in kinetic energy is ΔKE = ½m(v_f² - v_i²) = ½(5)((10)² - (4)²) = 2.5(100 - 16) = 2.5(84) = 210 J. Positive ΔKE means the object gained kinetic energy (sped up), while negative ΔKE means it lost kinetic energy (slowed down). Choice A is correct because it correctly calculates KE = ½mv² with velocity squared and ½ factor. Choice C forgets the ½ factor in the kinetic energy formula, calculating KE = mv² instead of KE = ½mv², which makes the kinetic energy twice what it should be. Remember that only net work determines kinetic energy change—individual forces can do positive or negative work, but it's the sum that matters.