A sled is pushed across a rough floor for 5.0m at constant speed by a horizontal force of 40N. Using the work–energy theorem, which statement is most consistent with the work done by friction during the motion?
Answer: Friction does negative work equal in magnitude to the applied force's positive work.
This question tests understanding of work done by friction using the work-energy theorem. At constant speed, kinetic energy is unchanged, so net work is zero. The applied force does positive work: W_applied = (40 N)(5.0 m) = 200 J. Since net work is zero, friction must do negative work equal in magnitude: W_friction = -200 J. The negative sign indicates friction opposes motion. Choice B incorrectly assumes constant speed means no forces act, when actually balanced forces act. Choice C incorrectly assigns positive work to a force opposing motion. When analyzing work at constant speed, remember that positive and negative work contributions must sum to zero.
In a muscle physiology assay, a frog muscle accelerates a 0.20kg lever from rest to 1.5m/s while moving it horizontally with negligible friction and no height change. Principle tested: work-energy theorem. Which statement is most consistent with the net work done by the muscle on the lever during the acceleration phase?
Answer: Net work equals the change in kinetic energy, ΔK=21mv2
The skill being tested is applying the work-energy theorem to accelerating systems. The work-energy theorem states that the net work done on an object equals its change in kinetic energy. In this muscle physiology assay, the lever is accelerated horizontally from rest with negligible friction and no potential energy change, so net work by the muscle equals the kinetic energy gained. Choice B is consistent because net work = ΔK = (1/2)mv², matching the theorem for the acceleration phase. Choice D is incorrect because it claims zero work due to horizontal motion, misunderstanding that work can change kinetic energy without height changes. For similar problems, calculate ΔK and set it equal to net work. Confirm by ensuring non-conservative forces like friction are negligible.
A biopsy needle is advanced 2.0cm into tissue while the clinician applies a nearly constant resistive force of 8N opposite the motion (assume the clinician's applied force matches this to maintain constant speed). Principle tested: sign of work. Which statement is most consistent with the work done by the resistive tissue force on the needle during insertion?
Answer: Negative work because the tissue force opposes the displacement
The skill being tested is determining the sign of work done by forces. The principle is that work is positive if force and displacement are in the same direction, negative if opposite. In this biopsy needle insertion, the tissue's resistive force opposes the displacement, so it does negative work on the needle. Choice B is consistent because the opposing force yields W = -F d. Choice C is incorrect because it claims zero work at constant speed, misunderstanding that individual forces do work even if net work is zero. In similar problems, assign sign based on force-displacement angle: cosθ <0 means negative. Calculate magnitude separately and add sign based on context.
In an isolated muscle-prep apparatus, a frog gastrocnemius lifts a 0.50kg mass vertically by 0.20m at constant speed. Neglect air resistance. Using energy conservation, which outcome is most consistent with the transformation of energy in this lift? (Use g=9.8m/s2.)
Answer: The mass gains gravitational potential energy equal to the work done by the muscle on the mass.
This question tests understanding of energy conservation during work done against gravity. When lifting an object at constant speed, the net force is zero, so the applied force equals the weight (mg). The work done by the muscle equals force times distance, or mgh = (0.50 kg)(9.8 m/s²)(0.20 m) = 0.98 J. This work is converted entirely into gravitational potential energy since kinetic energy remains constant (constant speed means no change in kinetic energy). Choice B is incorrect because constant speed means zero acceleration, not increasing acceleration. To verify energy transformations in lifting problems, check that work input equals the change in potential energy when speed is constant.
A microfluidic pump lifts blood (density ρ=1060kg/m3) through a vertical height difference of 0.50m at a volumetric flow rate of 1.0mL/s with negligible kinetic energy change. Principle tested: energy conservation for steady flow. Which prediction is most consistent with the required mechanical power to overcome gravity? (Use g=9.8m/s2; 1mL=1×10−6m3.)
Answer: Power is proportional to ρghV˙, so doubling V˙ doubles required power
The skill being tested is applying energy conservation to calculate power in fluid pumping against gravity. The principle is that for steady flow with negligible kinetic energy change, the mechanical power required equals the rate of gravitational potential energy increase, P = ρ g h Ṅ, where Ṅ is volumetric flow rate. In this microfluidic pump setup, blood is lifted at a constant flow rate, so power is directly proportional to flow rate and height. Choice A is consistent because doubling Ṅ doubles the mass flow rate, thus doubling the power needed to overcome gravity. Choice B is incorrect because it suggests power inversely scales with h, misunderstanding that higher lifts require more work per unit mass. In similar fluid energy problems, express power as mass flow rate times g h. Verify by checking units: kg/s m/s² m gives watts.
A researcher compares two stair-climb trials for the same subject. Trial 1: climb 2.5m vertical rise in 20s. Trial 2: climb the same rise in 10s. Assume negligible change in kinetic energy at start/end. Principle tested: power vs work. Which statement is most consistent with these trials?
Answer: Both trials require the same work against gravity, but Trial 2 has greater average power
The skill being tested is distinguishing work from power in gravitational energy changes. The principle is that work against gravity depends only on height change (W = mgh), while power is W/t, so shorter time means higher power for fixed work. In these stair-climb trials, negligible KE change means work is the same for both, but Trial 2 has higher power due to halved time. Choice B is consistent because it recognizes equal work but greater power in Trial 2. Choice A is incorrect because it claims twice the work for halved time, misunderstanding that work is path-independent for gravity. In similar problems, compute work as mgh regardless of time, then power as work/time. Check for constant speed to confirm no KE contribution.
A researcher uses a hand-crank generator to power a small implantable pump prototype. In Trial 1, the crank delivers W=60J of mechanical work in t=10s. In Trial 2, the same total work is delivered in t=5s. The core principle is the relationship between power and work, P=W/t. Which prediction is most consistent with this principle?
Answer: Trial 2 has higher average power because delivering the same work in less time increases P.
This question tests the relationship between power, work, and time. Power is defined as the rate of doing work, P = W/t, measuring how quickly energy is transferred. In Trial 1, P₁ = 60 J / 10 s = 6 W, while in Trial 2, P₂ = 60 J / 5 s = 12 W. Since the same work is done in half the time in Trial 2, the average power doubles. Choice B incorrectly assumes power depends only on total work, ignoring the critical time factor. When comparing power outputs, always divide the work done by the time taken, remembering that faster completion of the same task requires higher power.
A 1.0kg lab weight is raised 0.50m at constant speed using two different pulley setups. Setup X requires a larger applied force but a shorter pull distance; Setup Y requires a smaller applied force but a longer pull distance. Neglect friction. By conservation of energy, which statement is most consistent?
Answer: Both setups require the same input work to raise the weight by the same height.
This question tests conservation of energy in mechanical advantage systems. By conservation of energy, the work input must equal the work output (neglecting friction). The work output is the increase in gravitational potential energy: W = mgh = (1.0 kg)(9.8 m/s²)(0.50 m) = 4.9 J. Regardless of the pulley configuration, this same amount of work must be input. Setup X requires larger force but shorter distance, while Setup Y requires smaller force but longer distance, but the product (work) remains constant. Choice B and C incorrectly focus on only force or distance without considering their product. In ideal mechanical systems, work input equals work output regardless of mechanical advantage.
A motor protein moves a cargo along a microtubule at constant speed while exerting a roughly constant forward force. If the cargo travels twice the distance in the same direction, apply the work relationshipW=Fd (force parallel to displacement). Which outcome is most consistent for the work done by the motor?
Answer: The work doubles.
This question tests the direct proportionality between work and distance for constant force. When force is constant and parallel to displacement, work is W = Fd. If distance doubles while force remains constant, work doubles: W' = F(2d) = 2Fd = 2W. Choice B incorrectly assumes constant speed affects the work calculation, when only force and distance matter. Choice C incorrectly suggests work decreases with increased distance. For constant force problems, work is directly proportional to distance traveled in the direction of the force.
In a biomechanics test, a subject performs two lifts of the same box: Lift A raises it 0.50m vertically; Lift B raises it 0.25m. Both start and end at rest, and losses are negligible. Principle tested: work against gravity. Which prediction is most consistent with the mechanical work done on the box by the subject?
Answer: Lift A requires twice the work of Lift B because W=mgh
The skill being tested is comparing work against gravity for different height changes. The principle is that work on the box equals mgh for vertical lifts starting/ending at rest with negligible losses, scaling with h. Here, Lift A with twice the height requires twice the work of Lift B. Choice A is consistent because WA = mg(0.50) = 2 * mg(0.25) = 2 WB. Choice B is incorrect because it claims same work for same mass, misunderstanding dependence on height. In similar lifting problems, compute W = mgh directly. Confirm neglect of losses and rest conditions for pure potential change.
During a vertical jump measurement, a 60kg subject leaves the ground with center-of-mass speed 3.0m/s upward. Air resistance is negligible. Principle tested: conservation of mechanical energy. Which prediction is most consistent with the maximum rise in center-of-mass height after takeoff? (Use g=9.8m/s2.)
Answer: Height is h=2gv2, independent of mass
The skill being tested is using conservation of mechanical energy to predict motion in gravitational fields. The principle is that in the absence of nonconservative forces, total mechanical energy (KE + PE) is conserved, so initial kinetic energy converts to potential energy at maximum height. In this vertical jump, air resistance is negligible, so the subject's initial KE = (1/2)mv² converts entirely to mgh at peak height, yielding h = v²/(2g). Choice B is consistent because height is independent of mass, depending only on initial speed and g. Choice A is incorrect because it claims height depends on mass, misunderstanding that g acts equally on all masses in energy conservation. For similar projectile problems, set initial KE equal to final PE and solve for h. Verify by plugging in numbers: h ≈ (3)²/(2*9.8) ≈ 0.46 m.
A 5.0kg box is lifted vertically at constant speed by a rope with tension approximately equal to its weight. The box rises 1.0m. Using the work done by a force, which statement is most consistent about the work done by gravity on the box during the lift? (Use g=9.8m/s2.)
Answer: Gravity does negative work of magnitude mgh.
This question tests understanding work done by gravity during upward motion. Gravity acts downward with magnitude mg, while displacement is upward by height h. Since force and displacement are in opposite directions, work by gravity is negative: W_gravity = -mgh = -(5.0 kg)(9.8 m/s²)(1.0 m) = -49 J. The negative sign indicates gravity opposes the motion. Choice B incorrectly assigns positive work when force opposes displacement. Choice C incorrectly assumes constant speed means zero work by individual forces. When calculating work by gravity, use negative work for upward motion and positive work for downward motion.
In a respiratory mechanics demo, a piston compresses a gas in a syringe by pushing with a constant force over a measured distance. The piston is pushed twice as far while maintaining the same force direction and magnitude. By the work relationship, which outcome is most consistent for the work done on the gas (ignoring heat transfer details)?
Answer: The work done on the gas doubles.
This question tests the work relationship for constant force. Work equals force times distance when force is constant and parallel to displacement: W = Fd. If distance doubles while force remains constant, work doubles: W' = F(2d) = 2Fd = 2W. Choice B incorrectly assumes work depends only on force. Choice C incorrectly suggests work decreases with increased distance. Choice D incorrectly relates work sign to volume change rather than force direction. For constant force, work is directly proportional to distance traveled in the direction of force.
In a treadmill study of gait rehabilitation, a 70-kg patient walks up a ramp of height h=1.5m at constant speed. Neglecting air resistance and assuming no net change in kinetic energy, apply conservation of energy to predict the minimum increase in the patient's gravitational potential energy (with g=9.8m/s2). Which outcome is most consistent with the principle?
Answer: The patient's gravitational potential energy increases by mgh because chemical energy is converted into potential energy.
This question tests the application of conservation of energy to determine changes in gravitational potential energy. When an object moves upward at constant speed, the work-energy theorem tells us that the net work is zero (no change in kinetic energy), but energy must be supplied to increase the gravitational potential energy by mgh. In this case, the patient's muscles convert chemical energy into gravitational potential energy as they walk up the ramp, increasing the patient's potential energy by mgh = (70 kg)(9.8 m/s²)(1.5 m) = 1029 J. The correct answer recognizes that gravitational potential energy increases by mgh when height increases by h. Choice B incorrectly suggests potential energy decreases when moving upward, contradicting the fundamental relationship between height and gravitational potential energy. To verify energy changes in vertical motion, check that ΔPE = mgΔh, where Δh is positive for upward motion and negative for downward motion.
In a cycling ergometer test, two subjects each perform 300J of mechanical work on the pedals. Subject 1 completes the work in 10s; Subject 2 completes the same work in 20s. Using the definition of powerP=W/t, which prediction is most consistent with the data?
Answer: Subject 1 has higher power output because the same work is completed in less time.
This question tests the definition of power as the rate of doing work, expressed as P = W/t. Power measures how quickly work is performed, making it crucial for understanding the intensity of physical activities. Subject 1 completes 300 J of work in 10 s, yielding P₁ = 300 J / 10 s = 30 W, while Subject 2 completes the same work in 20 s, yielding P₂ = 300 J / 20 s = 15 W. The correct answer recognizes that Subject 1 has higher power output because the same work is done in less time. Choice A incorrectly suggests that longer time increases power, when in fact power is inversely proportional to time for fixed work. To compare power outputs, always calculate P = W/t for each case, remembering that faster completion of the same work means higher power.
A 0.20kg ball is dropped from rest from height h and lands with speed v, neglecting air resistance. A second identical ball is dropped from height 2h. Using energy conservation, which prediction is most consistent for the second ball's impact speed?
Answer: It is 2v.
This question tests energy conservation in free fall. By conservation of energy, potential energy converts to kinetic energy: mgh = ½mv². For the first ball, v² = 2gh. For the second ball dropped from 2h, v'² = 2g(2h) = 4gh = 2(2gh) = 2v². Therefore, v' = √2 × v. Choice B incorrectly assumes speed doubles when height doubles, forgetting that kinetic energy (not speed) is proportional to height. Choice D incorrectly suggests speed is independent of drop height. When relating height to impact speed in free fall, remember that speed is proportional to the square root of height.
A lab technician pushes a cart up a short ramp, starting and ending at rest. Compared with a frictionless ramp, the technician reports needing to apply more force over the same displacement when the ramp surface is rough. Principle tested: nonconservative work and mechanical energy. Which statement is most consistent with the added work input?
Answer: Additional work is dissipated mainly as thermal energy due to friction
The skill being tested is accounting for nonconservative work in energy conservation. The principle is that added input work due to friction dissipates as thermal energy, since mechanical energy (KE+PE) change is the same but requires more total work. In this rough ramp push, starting/ending at rest means ΔKE=0, ΔPE=mgh, but extra work overcomes friction, becoming heat. Choice A is consistent because it identifies thermal dissipation of additional work. Choice B is incorrect because it claims extra work increases final KE, but the cart ends at rest (KE=0). For similar frictional problems, set total work = ΔME + |W_friction|, where W_friction is negative for mechanical energy. Verify start/end at rest for ΔKE=0.
A researcher compares two elevators lifting identical loads through the same vertical height. Elevator A lifts the load in half the time of Elevator B. Assume both start and end at rest. By the power–work relationship, which conclusion is most consistent?
Answer: Elevator A has twice the average power output of Elevator B.
This question tests the relationship between power, work, and time. Both elevators do the same work (lifting identical loads through same height), but Elevator A completes it in half the time. Since power = work/time, and Elevator A has half the time for the same work, it has twice the power: PA = W/(t/2) = 2W/t = 2P_B. Choice B incorrectly relates work to speed rather than to force and distance. Choice C reverses the power-time relationship. When comparing power for equal work tasks, the faster completion time corresponds to higher power output.
A wearable metabolic device estimates that a cyclist produces the same mechanical work on the pedals during two 30 s intervals, but the cadence is higher in Interval 2 (more revolutions per second). The core principle is power as work per unit time. Which conclusion is most consistent if the intervals are both 30 s and total work is the same?
Answer: Average power is the same in both intervals because W and t are the same, regardless of cadence.
This question tests the concept of average power when work and time are held constant. Power is P = W/t, depending only on the total work done and the time taken. Since both intervals last 30 s and produce the same total mechanical work, the average power must be identical: P₁ = W/30s = P₂. The higher cadence in Interval 2 means more revolutions but less work per revolution. Choice B incorrectly assumes higher cadence automatically means more work, ignoring the given constraint of equal total work. When comparing power outputs, focus on the total work and time—the specific pattern of force application (cadence) doesn't matter if these totals are fixed.
In a biomechanics test, a subject jumps vertically. Just after takeoff, the subject's center of mass has speed v and then rises an additional height h before momentarily coming to rest. Ignore air resistance. The core principle tested is conservation of mechanical energy (21mv2=mgh during the rise). Which prediction is most consistent with this principle if the takeoff speed is doubled (from v to 2v)?
Answer: The additional rise height becomes four times larger because kinetic energy scales with v2.
This question tests conservation of mechanical energy, specifically how kinetic energy converts to gravitational potential energy. During vertical motion without air resistance, initial kinetic energy equals final potential energy: ½mv² = mgh. If takeoff speed doubles from v to 2v, kinetic energy increases by a factor of 4 (since KE ∝ v²): ½m(2v)² = 4(½mv²). Since this kinetic energy converts entirely to potential energy mgh', the new height h' = 4h. Choice B incorrectly assumes linear scaling, while choice D invents a nonexistent speed-dependent gravity effect. To solve energy conversion problems, remember that kinetic energy scales with velocity squared, leading to quadratic relationships in energy transformations.
