Physics Quiz: Apply Work Energy Theorem
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Apply Work Energy TheoremQuestion 1 of 20

A 5.0 kg5.0\,\text{kg} cart speeds up from 4.0 m/s4.0\,\text{m/s} to 10.0 m/s10.0\,\text{m/s} while moving along a straight horizontal track. What is the change in the cart's kinetic energy, ΔKE=KEf−KEi\Delta KE = KE_f-KE_i?

210 J210\,\text{J}
170 J170\,\text{J}
420 J420\,\text{J}
−210 J-210\,\text{J}
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Physics Quiz: Apply Work Energy Theorem

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

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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.

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Question 1

A 5.0 kg5.0\,\text{kg} cart speeds up from 4.0 m/s4.0\,\text{m/s} to 10.0 m/s10.0\,\text{m/s} while moving along a straight horizontal track. What is the change in the cart's kinetic energy, ΔKE=KEf−KEi\Delta KE = KE_f-KE_i?

  1. 210 J210\,\text{J} (correct answer)
  2. 170 J170\,\text{J}
  3. 420 J420\,\text{J}
  4. −210 J-210\,\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 = 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.

Question 2

A 5.0 kg5.0\,\text{kg} sled slides down a frictionless incline and speeds up from 2.0 m/s2.0\,\text{m/s} to 10.0 m/s10.0\,\text{m/s}. What is the net work done on the sled during this motion? (Include sign.)

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

A 4.0 kg4.0\,\text{kg} cart speeds up from 3.0 m/s3.0\,\text{m/s} to 9.0 m/s9.0\,\text{m/s} on a level track. During this process, what is the change in the cart's kinetic energy, ΔKE=KEf−KEi\Delta KE = KE_f-KE_i?

  1. 144 J144\,\text{J} (correct answer)
  2. 72 J72\,\text{J}
  3. 162 J162\,\text{J}
  4. 324 J324\,\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 4

A 20 kg20\,\text{kg} crate is pulled across a frictionless horizontal floor by a 100 N100\,\text{N} force applied at an angle of 30∘30^\circ above the horizontal. The crate moves 6.0 m6.0\,\text{m} horizontally. What is the work done by the applied force on the crate?

  1. 600 J600\,\text{J}
  2. 520 J520\,\text{J} (correct answer)
  3. 300 J300\,\text{J}
  4. 0 J0\,\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 5

A 1200 kg1200\,\text{kg} car traveling at 22 m/s22\,\text{m/s} brakes to a stop over a distance of 55 m55\,\text{m} on level ground. Assuming the braking force is constant and is the only horizontal force doing work, what is the magnitude of the braking force?

  1. 5.3×103 N5.3\times 10^3\,\text{N} (correct answer)
  2. 1.1×104 N1.1\times 10^4\,\text{N}
  3. 2.6×103 N2.6\times 10^3\,\text{N}
  4. 5.8×102 N5.8\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 6

A 10 kg10\,\text{kg} box slides 12 m12\,\text{m} across a rough horizontal floor. The only horizontal force doing work is kinetic friction with constant magnitude 15 N15\,\text{N} opposing the motion. What is the work done by friction on the box (include sign)?

  1. −180 J-180\,\text{J} (correct answer)
  2. 180 J180\,\text{J}
  3. −1.25 J-1.25\,\text{J}
  4. −360 J-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 7

A 2.0 kg2.0\,\text{kg} object is lifted straight upward 3.0 m3.0\,\text{m} at constant speed. Neglect air resistance. What is the net work done on the object during the lift?

  1. 58.8 J58.8\,\text{J}
  2. 0 J0\,\text{J} (correct answer)
  3. −58.8 J-58.8\,\text{J}
  4. 19.6 J19.6\,\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 8

A 10 kg10\,\text{kg} sled moves 8.0 m8.0\,\text{m} to the right on level snow. A person pulls with a constant force of 60 N60\,\text{N} at an angle of 30∘30^\circ above the horizontal. Kinetic friction is 15 N15\,\text{N} opposing the motion. What is the net work done on the sled over the 8.0 m8.0\,\text{m} displacement?

  1. 600 J600\,\text{J}
  2. 296 J296\,\text{J} (correct answer)
  3. 536 J536\,\text{J}
  4. −296 J-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 9

A 30 kg30\,\text{kg} suitcase is pulled 12 m12\,\text{m} across a level airport floor. The pulling force is 70 N70\,\text{N} at 0∘0^\circ (horizontal), and kinetic friction is 25 N25\,\text{N} opposing the motion. How does the work done by friction compare to the work done by the applied force (ratio Wfriction/WappliedW_{\text{friction}}/W_{\text{applied}})?

  1. −0.36-0.36 (correct answer)
  2. +0.36+0.36
  3. −2.8-2.8
  4. +2.8+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 10

A 1200 kg1200\,\text{kg} car is traveling at 24 m/s24\,\text{m/s} and brakes to a stop over a distance of 48 m48\,\text{m} on level ground. Assume the braking force is constant and is the only horizontal force doing work. What is the magnitude of the braking force? (Use Wnet=ΔKEW_{\text{net}}=\Delta KE.)

  1. 7.2×103 N7.2\times 10^3\,\text{N} (correct answer)
  2. 1.44×104 N1.44\times 10^4\,\text{N}
  3. 2.88×104 N2.88\times 10^4\,\text{N}
  4. 3.6×103 N3.6\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 11

A 10 kg10\,\text{kg} box is pulled across a horizontal floor for 8.0 m8.0\,\text{m} by a constant force of 60 N60\,\text{N} at an angle of 30∘30^\circ above the horizontal. A constant friction force of 20 N20\,\text{N} opposes the motion. What is the net work done on the box (in J) over the 8.0 m8.0\,\text{m} displacement? (Use W=Fdcos⁡θW = Fd\cos\theta for the applied force.)

  1. 256 J256\,\text{J} (correct answer)
  2. 416 J416\,\text{J}
  3. 96 J96\,\text{J}
  4. −256 J-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 12

A 18 kg18\,\text{kg} box slides on a horizontal surface with initial speed 10 m/s10\,\text{m/s}. Kinetic friction is constant at 45 N45\,\text{N} and is the only horizontal force doing work. Using Wnet=ΔKEW_{\text{net}}=\Delta KE with vf=0v_f=0, what stopping distance dd (in m) is required for the box to come to rest?

  1. 40 m40\,\text{m}
  2. 20 m20\,\text{m} (correct answer)
  3. 10 m10\,\text{m}
  4. 5.0 m5.0\,\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 13

A 6.0 kg6.0\,\text{kg} crate slides 10 m10\,\text{m} across a horizontal floor at constant speed while a person pulls it with a 40 N40\,\text{N} horizontal force. In this process, what is the work done by friction on the crate (include sign, in J)?

  1. +400 J+400\,\text{J}
  2. −400 J-400\,\text{J} (correct answer)
  3. 0 J0\,\text{J}
  4. −200 J-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 14

A 20 kg20\,\text{kg} cart speeds up from 3.0 m/s3.0\,\text{m/s} to 9.0 m/s9.0\,\text{m/s} on a level track. During this process, what is the change in the cart's kinetic energy ΔKE\Delta KE (in J)?

  1. 720 J720\,\text{J} (correct answer)
  2. 1620 J1620\,\text{J}
  3. 810 J810\,\text{J}
  4. 1440 J1440\,\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 15

A 25 kg25\,\text{kg} suitcase is pulled 5.0 m5.0\,\text{m} across a horizontal floor at constant speed. The pulling force is 100 N100\,\text{N} directed 37∘37^\circ above the horizontal. In this process, what is the work done by the pulling force on the suitcase? (Use W=Fdcos⁡θW = Fd\cos\theta.)

  1. 500 J500\,\text{J}
  2. 400 J400\,\text{J} (correct answer)
  3. 300 J300\,\text{J}
  4. −400 J-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 16

A 8.0 kg8.0\,\text{kg} cart moves on a level surface. A constant applied force of 50 N50\,\text{N} acts in the direction of motion while kinetic friction of 18 N18\,\text{N} opposes the motion. Over a displacement of 7.0 m7.0\,\text{m}, what is the net work done on the cart (in J)?

  1. 476 J476\,\text{J}
  2. 224 J224\,\text{J} (correct answer)
  3. 350 J350\,\text{J}
  4. 126 J126\,\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 17

A 30 kg30\,\text{kg} crate is pushed 5.0 m5.0\,\text{m} across a horizontal floor. The applied force is 200 N200\,\text{N} to the right. Kinetic friction is 50 N50\,\text{N} to the left. During this process, which expression correctly applies the work-energy theorem to relate the forces and displacement to the change in speed?

  1. (200 N)(5.0 m)+(50 N)(5.0 m)=12(30 kg)(vf2−vi2)(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 N)(5.0 m)−(50 N)(5.0 m)=12(30 kg)(vf2−vi2)(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 N)/(5.0 m)−(50 N)/(5.0 m)=12(30 kg)(vf2−vi2)(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 N)(5.0 m)−(50 N)(5.0 m)=(30 kg)(vf−vi)(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 18

A 25 kg25\,\text{kg} cart moving at 8.0 m/s8.0\,\text{m/s} is brought to 2.0 m/s2.0\,\text{m/s} by a constant friction force over a distance of 10 m10\,\text{m} on level ground. What is the magnitude of the friction force? (Take friction as the only horizontal force doing work.)

  1. 75 N75\,\text{N} (correct answer)
  2. 45 N45\,\text{N}
  3. −75 N-75\,\text{N}
  4. 150 N150\,\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 19

A 1200 kg1200\,\text{kg} car traveling at 24 m/s24\,\text{m/s} brakes to a stop over a distance of 48 m48\,\text{m} on level ground. Assume the braking force is constant and is the only horizontal force doing work. Using Wnet=ΔKEW_{\text{net}}=\Delta KE, what is the magnitude of the braking force (in N)?

  1. 1.5×104 N1.5\times 10^4\,\text{N}
  2. 7.2×103 N7.2\times 10^3\,\text{N} (correct answer)
  3. 3.6×103 N3.6\times 10^3\,\text{N}
  4. 2.9×105 N2.9\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 20

A 2.5 kg2.5\,\text{kg} object is pulled along a horizontal surface. A constant 40 N40\,\text{N} force acts in the direction of motion while a constant 10 N10\,\text{N} friction force acts opposite the motion. Over a displacement of 4.0 m4.0\,\text{m}, what is the net work done on the object?

  1. 160 J160\,\text{J}
  2. 120 J120\,\text{J} (correct answer)
  3. −120 J-120\,\text{J}
  4. 30 J30\,\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).