Physics Quiz: Apply Work Energy Theorem

What this quiz covers

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

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. $144\,\text{J}$ (correct answer)
2. $72\,\text{J}$
3. $162\,\text{J}$
4. $324\,\text{J}$

Explanation: 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).

Question 2

1. $600\,\text{J}$
2. $520\,\text{J}$ (correct answer)
3. $300\,\text{J}$
4. $0\,\text{J}$

Explanation: 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).

Question 3

1. $5.3\times 10^3\,\text{N}$ (correct answer)
2. $1.1\times 10^4\,\text{N}$
3. $2.6\times 10^3\,\text{N}$
4. $5.8\times 10^2\,\text{N}$

Explanation: 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).

Question 4

1. $-180\,\text{J}$ (correct answer)
2. $180\,\text{J}$
3. $-1.25\,\text{J}$
4. $-360\,\text{J}$

Explanation: 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.

Question 5

1. $58.8\,\text{J}$
2. $0\,\text{J}$ (correct answer)
3. $-58.8\,\text{J}$
4. $19.6\,\text{J}$

Explanation: 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 work: In this scenario, the object is lifted at constant speed, so net force is zero, implying W_net = 0 J. The work by lifting force is positive, W_lift = m g d ≈ 2 × 9.8 × 3 = 58.8 J, but work by gravity is W_gravity = -m g d ≈ -58.8 J, so W_net = 58.8 + (-58.8) = 0 J. Since constant speed, ΔKE = 0, consistent with W_net = ΔKE. Choice B is correct because it correctly accounts for sign of work: positive for forces in direction of motion, negative for opposing forces. Choice A reports the work done by individual force when the question asks for net work, which requires summing the work done by all forces: W_net = W_applied + W_friction + W_gravity. 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.

Question 6

1. $600\,\text{J}$
2. $296\,\text{J}$ (correct answer)
3. $536\,\text{J}$
4. $-296\,\text{J}$

Explanation: 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 work: In this scenario, applied force of magnitude F = 60 N acts over displacement d = 8.0 m at angle θ = 30° above horizontal. The work done is W = Fd cos(θ) = (60)(8) cos(30) = 480 × (√3/2) ≈ 480 × 0.866 = 415.7 J. If multiple forces, calculate each: W_net = W_applied + W_friction + W_gravity = 415.7 + (-120) + 0 = 295.7 J ≈ 296 J. Choice B is correct because it correctly accounts for sign of work: positive for forces in direction of motion, negative for opposing forces. Choice C 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. 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.

Question 7

1. $-0.36$ (correct answer)
2. $+0.36$
3. $-2.8$
4. $+2.8$

Explanation: 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 work: In this scenario, applied force of magnitude F = 70 N acts over displacement d = 12 m in same direction. The work done is W_applied = F d = 70 × 12 = 840 J, and W_friction = -25 × 12 = -300 J. The ratio W_friction / W_applied = -300 / 840 ≈ -0.36. Choice A is correct because it correctly accounts for sign of work: positive for forces in direction of motion, negative for opposing forces. 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. 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).

Question 8

1. $7.2\times 10^3\,\text{N}$ (correct answer)
2. $1.44\times 10^4\,\text{N}$
3. $2.88\times 10^4\,\text{N}$
4. $3.6\times 10^3\,\text{N}$

Explanation: 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 = 24 m/s over distance d = 48 m, we use W_net = ΔKE. The change in kinetic energy is ΔKE = 0 - ½m(v_i²) = -½(1200)(24)² = -600 × 576 = -345600 J (negative because kinetic energy decreases). Since W_net = -F × d and W_net = ΔKE, we have -F × (48) = -345600, giving F = 345600/48 = 7200 N in magnitude (negative sign indicates force opposes motion). Choice A is correct because it accurately applies work-energy theorem W_net = ½m(v_f² - v_i²) to solve for unknown. Choice B 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. 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).

Question 9

1. $256\,\text{J}$ (correct answer)
2. $416\,\text{J}$
3. $96\,\text{J}$
4. $-256\,\text{J}$

Explanation: 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, an applied force of 60 N acts at angle θ = 30° above horizontal over displacement d = 8.0 m, while friction of 20 N opposes motion. The work done by the applied force is W_applied = Fd cos(θ) = (60)(8.0) cos(30°) = (60)(8.0)(0.866) = 415.7 J, and the work done by friction is W_friction = -fd = -(20)(8.0) = -160 J (negative because it opposes motion). The net work is W_net = W_applied + W_friction = 415.7 + (-160) = 255.7 ≈ 256 J. Choice A is correct because it properly calculates the net work as approximately 256 J, accounting for both the angled applied force and the opposing friction force. Choice B (416 J) reports the work done by the applied force alone when the question asks for net work, which requires summing the work done by all forces: W_net = W_applied + W_friction. 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).

Question 10

1. $40\,\text{m}$
2. $20\,\text{m}$ (correct answer)
3. $10\,\text{m}$
4. $5.0\,\text{m}$

Explanation: 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 box (v_f = 0) from initial velocity v_i = 10 m/s, we use W_net = ΔKE. The change in kinetic energy is ΔKE = 0 - ½m(v_i²) = -½(18 kg)(10 m/s)² = -½(18)(100) = -900 J (negative because kinetic energy decreases). Since W_net = F_friction × d and W_net = ΔKE, we have (-45 N) × d = -900 J (friction opposes motion, so force is negative), giving d = -900 J / (-45 N) = 20 m. Choice B is correct because it properly applies the work-energy theorem to find the stopping distance, accounting for the negative work done by friction that removes all the box's kinetic energy. Choice A (40 m) likely uses velocity instead of velocity squared in the kinetic energy formula, calculating ½mv instead of ½mv², which dramatically underestimates the initial kinetic energy and overestimates the stopping distance. When finding stopping distance: (1) calculate initial kinetic energy KE_i = ½mv_i², (2) recognize that all this energy must be removed by friction work, so W_friction = -KE_i, and (3) use W = Fd to solve for distance. Key insight: the work-energy theorem directly relates stopping distance to initial speed squared—doubling speed quadruples stopping distance.

Question 11

1. $+400\,\text{J}$
2. $-400\,\text{J}$ (correct answer)
3. $0\,\text{J}$
4. $-200\,\text{J}$

Explanation: 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. In this scenario, the crate moves at constant speed, meaning v_f = v_i, so ΔKE = 0. By the work-energy theorem, W_net = 0, which means W_applied + W_friction = 0. The applied force does work W_applied = (40 N)(10 m) = +400 J (positive because force is in direction of motion). Since W_applied + W_friction = 0, we have 400 J + W_friction = 0, giving W_friction = -400 J. Choice B is correct because friction opposes motion, so its work is negative: W_friction = -f·d = -400 J, which removes kinetic energy at the same rate the applied force adds it, maintaining constant speed. Choice A (+400 J) incorrectly treats friction work as positive—friction opposes motion so its work must be negative to balance the positive work done by the applied force. When objects move at constant speed on a horizontal surface: (1) kinetic energy doesn't change, so ΔKE = 0, (2) by work-energy theorem, net work is zero: W_net = 0, (3) this means positive work by applied forces exactly cancels negative work by friction: W_applied + W_friction = 0.

Question 12

1. $720\,\text{J}$ (correct answer)
2. $1620\,\text{J}$
3. $810\,\text{J}$
4. $1440\,\text{J}$

Explanation: 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 kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). The cart has mass m = 20 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(20)((9.0)² - (3.0)²) = ½(20)(81 - 9) = ½(20)(72) = 720 J. Choice A is correct because it properly calculates ΔKE = ½(20)(81 - 9) = ½(20)(72) = 720 J, showing the cart gained kinetic energy as it sped up. Choice D (1440 J) forgets the ½ factor in the kinetic energy formula, calculating KE = mv² instead of KE = ½mv², which makes the kinetic energy change twice what it should be: (20)(72) = 1440 J instead of 720 J. When calculating kinetic energy changes: (1) always include the ½ factor in KE = ½mv², (2) square the velocities before subtracting: v_f² - v_i², not (v_f - v_i)², (3) positive ΔKE means the object gained kinetic energy (sped up), while negative ΔKE means it lost kinetic energy (slowed down).

Question 13

1. $500\,\text{J}$
2. $400\,\text{J}$ (correct answer)
3. $300\,\text{J}$
4. $-400\,\text{J}$

Explanation: 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 work is calculated as W = Fd cos(θ) if force at angle θ from displacement direction, where cos(θ) gives the component of force in the direction of motion. In this scenario, a pulling force of magnitude F = 100 N acts at angle θ = 37° above horizontal over displacement d = 5.0 m. The work done is W = Fd cos(θ) = (100)(5.0) cos(37°) = (100)(5.0)(0.8) = 400 J. Choice B is correct because it properly applies W = Fd cos(37°) = (100)(5.0)(0.8) = 400 J, using cos(37°) ≈ 0.8 to find the horizontal component of the angled force. Choice A (500 J) incorrectly calculates the work as if the force were horizontal, using W = Fd = (100)(5.0) = 500 J without accounting for the angle, which overestimates the work done. When calculating work by angled forces: (1) identify the angle θ between force and displacement directions, (2) use W = Fd cos(θ), not just W = Fd, (3) remember that only the component of force in the direction of motion (F cos(θ)) does work on the object.

Question 14

1. $476\,\text{J}$
2. $224\,\text{J}$ (correct answer)
3. $350\,\text{J}$
4. $126\,\text{J}$

Explanation: 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. In this scenario, an applied force of 50 N acts in the direction of motion while friction of 18 N opposes motion, both over displacement d = 7.0 m. The net work is W_net = W_applied + W_friction = (50)(7.0) + (-18)(7.0) = 350 + (-126) = 224 J. Choice B (224 J) is correct because it properly accounts for both the positive work by the applied force and the negative work by friction to find the net work. Choice A (476 J) reports the work done by the applied force alone (350 J) plus the friction work treated as positive (126 J), when friction work should be negative since it opposes motion. When calculating net work: sum the work done by all forces, remembering that forces in the direction of motion do positive work while opposing forces (like friction) do negative work, and it's the net work that determines kinetic energy change.

Question 15

1. $(200\,\text{N})(5.0\,\text{m}) + (50\,\text{N})(5.0\,\text{m}) = \tfrac{1}{2}(30\,\text{kg})\left(v_f^2 - v_i^2\right)$
2. $(200\,\text{N})(5.0\,\text{m}) - (50\,\text{N})(5.0\,\text{m}) = \tfrac{1}{2}(30\,\text{kg})\left(v_f^2 - v_i^2\right)$ (correct answer)
3. $(200\,\text{N})/(5.0\,\text{m}) - (50\,\text{N})/(5.0\,\text{m}) = \tfrac{1}{2}(30\,\text{kg})\left(v_f^2 - v_i^2\right)$
4. $(200\,\text{N})(5.0\,\text{m}) - (50\,\text{N})(5.0\,\text{m}) = (30\,\text{kg})\left(v_f - v_i\right)$

Explanation: 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 (positive if in direction of motion, negative if opposing). For the crate being pushed, the applied force of 200 N does positive work (in direction of motion) while friction of 50 N does negative work (opposes motion). The correct expression for net work is W_net = W_applied + W_friction = (200 N)(5.0 m) + (-(50 N)(5.0 m)) = (200 N)(5.0 m) - (50 N)(5.0 m), and this equals the change in kinetic energy ½m(v_f² - v_i²). Choice B correctly shows (200 N)(5.0 m) - (50 N)(5.0 m) = ½(30 kg)(v_f² - v_i²), properly subtracting friction work and using the kinetic energy formula with ½ factor and velocity squared. Choice A incorrectly adds the friction work instead of subtracting it—friction opposes motion so its work must be negative. Choice D uses (v_f - v_i) instead of (v_f² - v_i²) and omits the ½ factor, fundamentally misrepresenting the kinetic energy formula. When setting up work-energy equations: (1) express work by each force as force times displacement, (2) use positive sign for forces in direction of motion, negative for opposing forces, (3) set net work equal to ½m(v_f² - v_i²) with proper ½ factor and squared velocities.

Question 16

1. $75\,\text{N}$ (correct answer)
2. $45\,\text{N}$
3. $-75\,\text{N}$
4. $150\,\text{N}$

Explanation: 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 slow an object from v_i = 8.0 m/s to v_f = 2.0 m/s over distance d = 10 m, we use W_net = ΔKE. The change in kinetic energy is ΔKE = ½m(v_f² - v_i²) = ½(25)(4 - 64) = 12.5 × (-60) = -750 J (negative because kinetic energy decreases). Since W_net = -f × d and W_net = ΔKE, we have -f × (10) = -750, giving f = 750/10 = 75 N in magnitude (negative sign indicates force opposes motion). Choice A is correct because it accurately applies work-energy theorem W_net = ½m(v_f² - v_i²) to solve for unknown. Choice D 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. 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.

Question 17

1. $+240\,\text{J}$ (correct answer)
2. $+120\,\text{J}$
3. $-240\,\text{J}$
4. $+480\,\text{J}$

Explanation: 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 kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). The sled has mass m = 5.0 kg, initial velocity v_i = 2.0 m/s, and final velocity v_f = 10.0 m/s. The net work done is W_net = ΔKE = ½m(v_f² - v_i²) = ½(5.0 kg)((10.0 m/s)² - (2.0 m/s)²) = ½(5.0)(100 - 4) = ½(5.0)(96) = 240 J. Since the sled speeds up (v_f > v_i), the net work is positive (+240 J), indicating that energy was added to the system. Choice A (+240 J) is correct because it properly applies the work-energy theorem with the correct ½ factor and velocity squared terms. Choice B (+120 J) is incorrect—it appears to result from forgetting to square the velocities, calculating something like ½m(v_f - v_i) × (v_f + v_i) = ½(5.0)(8)(12) = 240 J but then dividing by 2 again. When calculating net work from velocity changes: (1) calculate initial kinetic energy: KE_i = ½mv_i², (2) calculate final kinetic energy: KE_f = ½mv_f², (3) find net work as W_net = KE_f - KE_i, which is positive if the object speeds up and negative if it slows down.

Question 18

1. $1.5\times 10^4\,\text{N}$
2. $7.2\times 10^3\,\text{N}$ (correct answer)
3. $3.6\times 10^3\,\text{N}$
4. $2.9\times 10^5\,\text{N}$

Explanation: 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 a car (v_f = 0) from initial velocity v_i = 24 m/s over distance d = 48 m, we use W_net = ΔKE. The change in kinetic energy is ΔKE = 0 - ½m(v_i²) = -½(1200 kg)(24 m/s)² = -½(1200)(576) = -345,600 J (negative because kinetic energy decreases). Since W_net = F_braking × d and W_net = ΔKE, we have F_braking × (48 m) = -345,600 J, giving F_braking = -345,600 J / 48 m = -7,200 N in magnitude (negative sign indicates force opposes motion), so the magnitude is 7,200 N = 7.2 × 10³ N. Choice B is correct because it properly applies the work-energy theorem to find the braking force needed to bring the car to rest, accounting for the negative work done by the braking force. Choice D (2.9 × 10⁵ N) likely forgets the ½ factor in the kinetic energy formula, calculating KE = mv² instead of KE = ½mv², which makes the required force twice what it should be. When solving for forces using work-energy: (1) calculate the change in kinetic energy (negative for slowing down), (2) use W_net = F·d to find the force, remembering that braking forces do negative work, and (3) report magnitude when asked. Key insight: the work-energy theorem provides a powerful alternative to kinematics when dealing with problems involving force, displacement, and velocity changes.

Question 19

1. $160\,\text{J}$
2. $120\,\text{J}$ (correct answer)
3. $-120\,\text{J}$
4. $30\,\text{J}$

Explanation: 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).

Question 20

1. $3.75\times 10^3\,\text{N}$
2. $7.50\times 10^3\,\text{N}$ (correct answer)
3. $1.50\times 10^4\,\text{N}$
4. $7.50\times 10^4\,\text{N}$

Explanation: 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). To stop an object (v_f = 0) from initial velocity v_i = 25 m/s over distance d = 50 m, we use W_net = ΔKE. The change in kinetic energy is ΔKE = 0 - ½m(v_i²) = -½(1200)(25)² = -½(1200)(625) = -375,000 J (negative because kinetic energy decreases). Since W_net = F_braking × d and W_net = ΔKE, we have F_braking × (50) = -375,000 J, giving F_braking = -375,000/50 = -7,500 N in magnitude (negative sign indicates force opposes motion), which equals 7.50 × 10³ N. Choice B is correct because it properly applies the work-energy theorem to find the braking force magnitude of 7,500 N needed to bring the car to rest. Choice C (1.50 × 10⁴ N) 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 and thus doubles the required braking 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).