Apply Work-Energy Theorem - Physics

A $6.0\,\text{kg}$ crate slides $10\,\text{m}$ across a horizontal floor at constant speed while a person pulls it with a $40\,\text{N}$ horizontal force. In this process, what is the work done by friction on the crate (include sign, in J)?
Answer: $-400\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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.
A $30\,\text{kg}$ box is pulled $5.0\,\text{m}$ across a level floor at constant speed. The pulling force is $70\,\text{N}$ directed horizontally. What is the work done by friction on the box (in J)? (Include sign.)
Answer: $-350\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), where work is calculated as W = Fd for a constant force parallel to displacement. Since the box moves at constant speed, its kinetic energy doesn't change, so ΔKE = 0, which means W_net = 0. The pulling force does work W_pull = (70 N)(5.0 m) = +350 J (positive because force is in direction of motion). Since W_net = W_pull + W_friction = 0, we have 350 J + W_friction = 0, giving W_friction = -350 J. Choice B (-350 J) is correct because friction opposes motion so its work is negative, and the magnitude equals the positive work done by the pulling force to maintain constant speed. Choice A (+350 J) incorrectly treats the friction work as positive instead of negative—friction opposes motion so its work is W_friction = -f·d (negative), removing energy that the pulling force adds. When an object moves at constant speed on a level surface, the net work is zero because kinetic energy remains constant, and opposing forces (like friction) do negative work equal in magnitude to the positive work done by applied forces.
A $30\,\text{kg}$ suitcase is pulled $12\,\text{m}$ across a level airport floor. The pulling force is $70\,\text{N}$ at $0^\circ$ (horizontal), and kinetic friction is $25\,\text{N}$ opposing the motion. How does the work done by friction compare to the work done by the applied force (ratio $W_{\text{friction}}/W_{\text{applied}}$ )?
Answer: $-0.36$ 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( $v_f$ ² - $v_i$ ²), 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).
A $7.0\,\text{kg}$ cart moves on a level track with initial speed $10\,\text{m/s}$ . A constant net force opposite the motion acts over $14\,\text{m}$ , doing $-350\,\text{J}$ of net work on the cart. Using $W_{\text{net}}=\Delta KE$ , what is the cart's final speed?
Answer: $0\,\text{m/s}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_f$ ² - $v_i$ ²) with m = 7.0 kg and $v_i$ = 10 m/s: -350 J = ½(7)( $v_f$ ² - 100), solving for $v_f$ : $v_f$ ² = 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).
A $2.0\,\text{kg}$ cart on a level track has an initial speed of $12\,\text{m/s}$ . A constant friction force of $4.0\,\text{N}$ acts opposite the motion until the cart stops. Using $W_{\text{net}}=\Delta KE$ , how far does the cart travel before stopping (in m)?
Answer: $36\,\text{m}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_i$ ²) = -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.
A $2.5\,\text{kg}$ object is pulled along a horizontal surface. A constant $40\,\text{N}$ force acts in the direction of motion while a constant $10\,\text{N}$ friction force acts opposite the motion. Over a displacement of $4.0\,\text{m}$ , what is the net work done on the object?
Answer: $120\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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).
A $10\,\text{kg}$ box slides $6.0\,\text{m}$ to the right on a horizontal floor. A kinetic friction force of $25\,\text{N}$ acts to the left, and a horizontal applied force of $40\,\text{N}$ acts to the right. What is the net work done on the box over the $6.0\,\text{m}$ displacement? (Include sign.)
Answer: $+90\,\text{J}$ 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.
A $4.0\,\text{kg}$ cart speeds up from $3.0\,\text{m/s}$ to $9.0\,\text{m/s}$ on a level track. During this process, what is the change in the cart's kinetic energy, $\Delta KE = KE_f-KE_i$ ?
Answer: $144\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_f$ ² - $v_i$ ²) = 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( $v_f$ ² - $v_i$ ²)/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).
A $10\,\text{kg}$ box is pulled across a horizontal floor for $8.0\,\text{m}$ by a constant force of $60\,\text{N}$ at an angle of $30^\circ$ above the horizontal. A constant friction force of $20\,\text{N}$ opposes the motion. What is the net work done on the box (in J) over the $8.0\,\text{m}$ displacement? (Use $W = Fd\cos\theta$ for the applied force.)
Answer: $256\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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).
A $25\,\text{kg}$ suitcase is pulled $5.0\,\text{m}$ across a horizontal floor at constant speed. The pulling force is $100\,\text{N}$ directed $37^\circ$ above the horizontal. In this process, what is the work done by the pulling force on the suitcase? (Use $W = Fd\cos\theta$ .)
Answer: $400\,\text{J}$ 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.
A $2.0\,\text{kg}$ object is lifted straight upward $3.0\,\text{m}$ at constant speed. Neglect air resistance. What is the net work done on the object during the lift?
Answer: $0\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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.
A $10\,\text{kg}$ box slides $12\,\text{m}$ across a rough horizontal floor. The only horizontal force doing work is kinetic friction with constant magnitude $15\,\text{N}$ opposing the motion. What is the work done by friction on the box (include sign)?
Answer: $-180\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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.
A $30\,\text{kg}$ crate is pushed $5.0\,\text{m}$ across a horizontal floor. The applied force is $200\,\text{N}$ to the right. Kinetic friction is $50\,\text{N}$ 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?
Answer: $(200\,\text{N})(5.0\,\text{m}) - (50\,\text{N})(5.0\,\text{m}) = \tfrac{1}{2}(30\,\text{kg})\left(v_f^2 - v_i^2\right)$ 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( $v_f$ ² - $v_i$ ²), 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( $v_f$ ² - $v_i$ ²). Choice B correctly shows (200 N)(5.0 m) - (50 N)(5.0 m) = ½(30 kg)( $v_f$ ² - $v_i$ ²), 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 ( $v_f$ ² - $v_i$ ²) 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( $v_f$ ² - $v_i$ ²) with proper ½ factor and squared velocities.
A $1200\,\text{kg}$ car traveling at $24\,\text{m/s}$ brakes to a stop over a distance of $48\,\text{m}$ on level ground. Assume the braking force is constant and is the only horizontal force doing work. Using $W_{\text{net}}=\Delta KE$ , what is the magnitude of the braking force (in N)?
Answer: $7.2\times 10^3\,\text{N}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_i$ ²) = -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.
A $1200\,\text{kg}$ car traveling at $22\,\text{m/s}$ brakes to a stop over a distance of $55\,\text{m}$ 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?
Answer: $5.3\times 10^3\,\text{N}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_i$ ²) = -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).
A $8.0\,\text{kg}$ cart moves on a level surface. A constant applied force of $50\,\text{N}$ acts in the direction of motion while kinetic friction of $18\,\text{N}$ opposes the motion. Over a displacement of $7.0\,\text{m}$ , what is the net work done on the cart (in J)?
Answer: $224\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), 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.
A $40\,\text{kg}$ shopping cart speeds up from $1.0\,\text{m/s}$ to $5.0\,\text{m/s}$ while moving $10\,\text{m}$ on a level floor. Using the work-energy theorem, what is the net work done on the cart $W_{\text{net}}$ (in J)?
Answer: $480\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), where kinetic energy is KE = ½mv² (note the ½ factor and velocity squared). The cart has mass m = 40 kg, initial velocity $v_i$ = 1.0 m/s, and final velocity $v_f$ = 5.0 m/s. The change in kinetic energy is ΔKE = ½m( $v_f$ ² - $v_i$ ²) = ½(40 kg)((5.0 m/s)² - (1.0 m/s)²) = ½(40)(25 - 1) = ½(40)(24) = 480 J. Choice A is correct because it properly applies W_net = ΔKE = ½m( $v_f$ ² - $v_i$ ²) with the correct ½ factor and squared velocities to find the net work done. Choice C (1040 J) forgets the ½ factor in the kinetic energy formula, calculating KE = mv² instead of KE = ½mv², which makes the energy change twice what it should be. When calculating work from velocity change: (1) calculate initial and final kinetic energies using KE = ½mv² (don't forget the ½ and v²), (2) find the change ΔKE = KE_f - KE_i, and (3) by the work-energy theorem, this equals the net work done. Common errors to avoid: forgetting the ½ in KE = ½mv² (makes energy twice too large) and using v instead of v² (dramatically underestimates KE).
A $25\,\text{kg}$ cart moving at $8.0\,\text{m/s}$ is brought to $2.0\,\text{m/s}$ by a constant friction force over a distance of $10\,\text{m}$ on level ground. What is the magnitude of the friction force? (Take friction as the only horizontal force doing work.)
Answer: $75\,\text{N}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_f$ ² - $v_i$ ²) = 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( $v_f$ ² - $v_i$ ²) 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.
A $18\,\text{kg}$ box slides on a horizontal surface with initial speed $10\,\text{m/s}$ . Kinetic friction is constant at $45\,\text{N}$ and is the only horizontal force doing work. Using $W_{\text{net}}=\Delta KE$ with $v_f=0$ , what stopping distance $d$ (in m) is required for the box to come to rest?
Answer: $20\,\text{m}$ 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( $v_f$ ² - $v_i$ ²), 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( $v_i$ ²) = -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.
A $15\,\text{kg}$ suitcase is pulled $10\,\text{m}$ along a horizontal floor at constant speed. The pulling force is $40\,\text{N}$ directed horizontally (parallel to the displacement). What is the work done by kinetic friction on the suitcase (include sign) in Joules?
Answer: $-400\,\text{J}$ 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( $v_f$ ² - $v_i$ ²), where work is calculated as W = Fd for a constant force parallel to displacement, and kinetic energy is KE = ½mv². Since the suitcase moves at constant speed, its kinetic energy doesn't change, so ΔKE = 0 and therefore W_net = 0. The work done by the pulling force is W_pull = (40 N)(10 m) = +400 J. Since W_net = W_pull + W_friction = 0, we have 400 J + W_friction = 0, which gives W_friction = -400 J. Choice C is correct because friction opposes motion so its work is negative: W_friction = -f·d = -400 J, exactly balancing the positive work done by the pulling force to maintain constant speed. Choice A (+400 J) treats the friction work as positive instead of negative—friction opposes motion so its work is always negative, removing energy that would otherwise increase the object's speed. When an object moves at constant speed on a horizontal surface: (1) the net work is zero (W_net = 0), (2) the work done by friction exactly cancels the work done by the applied force, and (3) friction work is always negative. Key principle: at constant velocity, all forces balance and net work equals zero.

Start this quiz