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MCAT Chemical and Physical Foundations of Biological Systems

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4a Work Energy Power

Practice 4a Work Energy Power in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 4a Work Energy Power, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

In a biomechanics test, a subject performs two lifts of the same box: Lift A raises it 0.50 m0.50\ \text{m}0.50 m vertically; Lift B raises it 0.25 m0.25\ \text{m}0.25 m. 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?

  1. Lift A requires twice the work of Lift B because W=mghW = mghW=mgh
  2. Lift A requires the same work as Lift B because the box mass is unchanged
  3. Lift B requires more work because shorter distance implies greater force
  4. Work is determined by time, so the slower lift requires more work
Explanation: 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 W_A = mg(0.50) = 2 * mg(0.25) = 2 W_B. 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.

Question 2

A researcher pulls a sled carrying equipment across a lab floor. The applied force is constant, but the direction of motion reverses (the sled is pulled back along the same line). Principle tested: work sign and directionality. Which statement is most consistent with the work done by the same applied force magnitude during the return trip if the force is still directed forward (original direction)?

  1. Work becomes negative because force and displacement are opposite
  2. Work stays positive because force magnitude is unchanged
  3. Work becomes zero because the path retraces itself
  4. Work cannot be negative; energy is always positive
Explanation: The skill being tested is understanding the sign of work when direction reverses. The principle is that work's sign depends on the angle between force and displacement; if they oppose, work is negative. In this sled scenario, on return, if force is still forward but displacement is backward, they oppose, so work is negative. Choice A is consistent because it notes the opposition leading to negative work. Choice B is incorrect because it claims positive work for unchanged magnitude, misunderstanding directional dependence. For similar reversal problems, redefine displacement direction and recompute cosθ. Confirm force direction remains constant relative to original.

Question 3

A student claims that because a patient’s arm exerts a large force while holding a heavy suitcase motionless, the arm must be doing large mechanical work on the suitcase. Principle tested: work requires displacement. Which statement best evaluates the claim?

  1. Correct; work is large because force is large even with zero displacement
  2. Incorrect; mechanical work on the suitcase is approximately zero because displacement is zero
  3. Correct; work equals the weight of the suitcase regardless of motion
  4. Incorrect; work is negative because the force is upward
Explanation: The skill being tested is recognizing that work requires displacement in mechanics. The principle is that mechanical work W = F d cosθ is zero if d=0, even if force is applied. In this suitcase scenario, motionless holding means d=0, so work on suitcase is zero despite large force. Choice B is consistent because it correctly evaluates the claim as incorrect due to zero displacement. Choice A is incorrect because it supports large work with zero d, misunderstanding the definition requiring displacement. For similar static problems, check if displacement is zero to conclude zero work. Distinguish physiological effort from mechanical work.

Question 4

During a physical therapy session, a patient lifts a 4.0 kg4.0\ \text{kg}4.0 kg dumbbell vertically upward by 0.50 m0.50\ \text{m}0.50 m at constant speed. Neglecting air resistance and taking g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2, apply the work–energy theorem to the dumbbell. Which statement is most consistent with the net work on the dumbbell?

  1. The net work on the dumbbell is positive because the dumbbell is moving upward.
  2. The net work on the dumbbell is zero because its kinetic energy does not change at constant speed.
  3. The net work on the dumbbell equals mghmghmgh because net work always equals the change in potential energy.
  4. The net work on the dumbbell is mgmgmg because work and force are equivalent for vertical motion.
Explanation: This question tests the work-energy theorem applied to an object moving at constant speed, requiring careful analysis of net work versus individual work contributions. The work-energy theorem states that net work equals the change in kinetic energy (W_net = ΔK), and since the dumbbell moves at constant speed, its kinetic energy doesn't change, making the net work zero. While the patient does positive work mgh on the dumbbell and gravity does negative work -mgh, these cancel out to give zero net work. The correct answer properly applies the work-energy theorem to conclude that net work is zero when kinetic energy is constant. Choice A incorrectly confuses the direction of motion with the sign of net work, failing to recognize that upward motion at constant speed still means zero net work. To verify net work, always check the change in kinetic energy: if speed is constant, net work must be zero regardless of the path taken.

Question 5

A researcher compares two ways of delivering the same mechanical work to a tissue sample using a motorized indenter. Protocol A applies a constant force of 2 N2\ \text{N}2 N over 0.10 m0.10\ \text{m}0.10 m in 1.0 s1.0\ \text{s}1.0 s. Protocol B applies 1 N1\ \text{N}1 N over 0.20 m0.20\ \text{m}0.20 m in 2.0 s2.0\ \text{s}2.0 s. Using work W=FdW = FdW=Fd (force parallel to displacement) and power P=W/tP = W/tP=W/t, which prediction is most consistent?

  1. Both protocols deliver the same work and the same average power.
  2. Protocol A delivers greater work because it uses a larger force.
  3. Protocol B delivers greater average power because it lasts longer.
  4. Both protocols deliver the same work, but Protocol A has higher average power.
Explanation: This question tests the calculation of work and power for different force-displacement protocols, requiring application of both W = Fd and P = W/t. Work depends only on force and displacement (W = Fd when force is parallel to displacement), while power depends on how quickly that work is done (P = W/t). Protocol A delivers W_A = (2 N)(0.10 m) = 0.20 J in 1.0 s, giving P_A = 0.20 J / 1.0 s = 0.20 W, while Protocol B delivers W_B = (1 N)(0.20 m) = 0.20 J in 2.0 s, giving P_B = 0.20 J / 2.0 s = 0.10 W. The correct answer recognizes that both protocols deliver the same work but Protocol A has higher average power due to shorter duration. Choice C incorrectly suggests that longer duration increases power, when actually power is inversely proportional to time for fixed work. When comparing protocols, calculate both work (W = Fd) and power (P = W/t) separately to avoid confusing these related but distinct quantities.

Question 6

In a treadmill study of gait rehabilitation, a 70-kg patient walks up a ramp of height h=1.5 mh = 1.5\ \text{m}h=1.5 m 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=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2). Which outcome is most consistent with the principle?

  1. The patient’s gravitational potential energy increases by mghmghmgh because chemical energy is converted into potential energy.
  2. The patient’s gravitational potential energy decreases by mghmghmgh because the ramp provides an upward normal force.
  3. The patient’s gravitational potential energy is unchanged because constant speed implies zero work by muscles.
  4. The patient’s gravitational potential energy increases by mg/hmg/hmg/h because power is inversely related to height.
Explanation: 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.

Question 7

A researcher measures the mechanical work done by a subject’s quadriceps during a controlled knee extension. The net external torque produces an increase in the lower leg’s rotational kinetic energy by 12 J12\ \text{J}12 J over the movement (losses neglected). By the work–energy theorem, which statement is most consistent with the measurement?

  1. The net work done on the lower leg is 12 J12\ \text{J}12 J because Wnet=ΔKW_{\text{net}} = \Delta KWnet​=ΔK.
  2. The net work done on the lower leg is 0 J0\ \text{J}0 J because internal muscle forces cannot do work on the body.
  3. The net work done on the lower leg is 12 N12\ \text{N}12 N because work and force have the same units in biomechanics.
  4. The net work done on the lower leg is −12 J-12\ \text{J}−12 J because increasing kinetic energy requires negative work.
Explanation: This question tests understanding of the work-energy theorem, which states that the net work done on an object equals its change in kinetic energy. The work-energy theorem (W_net = ΔK) is a fundamental principle that applies to all types of motion, including rotational motion. Since the lower leg's rotational kinetic energy increases by 12 J, the net work done on it must be exactly 12 J. The correct answer properly applies W_net = ΔK to conclude that 12 J of net work was done. Choice C incorrectly confuses units, as work is measured in joules (J), not newtons (N), while choice D incorrectly suggests that positive changes in kinetic energy require negative work. When applying the work-energy theorem, always remember that positive net work increases kinetic energy, negative net work decreases it, and the magnitude of net work equals the magnitude of the kinetic energy change.

Question 8

A lab models blood flow using a pump that raises V=5.0 mLV = 5.0\ \text{mL}V=5.0 mL of fluid (density ρ=1000 kg/m3\rho = 1000\ \text{kg/m}^3ρ=1000 kg/m3) by a vertical height h=0.80 mh = 0.80\ \text{m}h=0.80 m at constant speed. Neglecting losses and using g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2, apply conservation of mechanical energy to identify the mechanical energy change of the fluid. Which statement is most consistent?

  1. The fluid’s mechanical energy increases by mghmghmgh because gravitational potential energy increases.
  2. The fluid’s mechanical energy decreases by mghmghmgh because the pump does negative work on the fluid.
  3. The fluid’s mechanical energy is unchanged because constant speed implies zero change in potential energy.
  4. The fluid’s mechanical energy increases by ρg/h\rho g/hρg/h because density replaces mass in potential energy.
Explanation: This question tests conservation of mechanical energy applied to fluid systems, where mechanical energy includes both kinetic and potential energy. When fluid is raised at constant speed, its kinetic energy remains unchanged, but its gravitational potential energy increases by mgh, where m = ρV is the fluid's mass. For this system, m = (1000 kg/m³)(5.0 × 10⁻⁶ m³) = 0.005 kg, so the mechanical energy increase is mgh = (0.005 kg)(9.8 m/s²)(0.80 m) = 0.0392 J. The correct answer recognizes that mechanical energy increases by mgh due to the increase in gravitational potential energy. Choice C incorrectly claims that constant speed implies no change in potential energy, confusing kinetic energy (which doesn't change) with potential energy (which does change with height). When analyzing energy changes in fluid systems, remember that mechanical energy includes both kinetic and potential components, and height changes always affect gravitational potential energy.

Question 9

A student pushes a cart carrying tissue samples up a ramp at constant speed. The applied force is parallel to the ramp and the cart moves a distance ddd along the ramp. The core principle tested is work by a constant force (W=Fdcos⁡θW = Fd\cos\thetaW=Fdcosθ). Which statement best matches the sign of the work done by the student’s applied force on the cart during the upward motion?

  1. Positive, because the applied force has a component in the direction of the cart’s displacement.
  2. Negative, because the cart moves upward against gravity so all forces do negative work.
  3. Zero, because constant speed implies zero work by any individual force.
  4. Zero, because work depends only on force magnitude and not on displacement direction.
Explanation: This question tests the calculation of work by a constant force using W = Fd cos θ, where θ is the angle between force and displacement. Work is positive when force has a component in the direction of displacement. Since the applied force is parallel to the ramp and the cart moves up the ramp in the same direction, θ = 0° and cos θ = 1, making the work positive: W = Fd. The student pushes in the direction of motion, doing positive work. Choice B incorrectly assumes all forces do negative work when moving against gravity, while choice C wrongly claims constant speed means zero work by individual forces (only net work is zero). To determine work's sign, check if force and displacement point in similar (positive) or opposite (negative) directions.

Question 10

A microfluidic actuator does W=4.0×10−3 JW = 4.0\times10^{-3}\ \text{J}W=4.0×10−3 J of work on a flexible membrane during one cycle. It runs at f=50 Hzf = 50\ \text{Hz}f=50 Hz (50 cycles per second). The core principle tested is power from work per cycle: P=W×fP = W\times fP=W×f. Which value is most consistent with the actuator’s average power output?

  1. 2.0×10−1 W2.0\times10^{-1}\ \text{W}2.0×10−1 W
  2. 8.0×10−5 W8.0\times10^{-5}\ \text{W}8.0×10−5 W
  3. 2.0×10−4 J2.0\times10^{-4}\ \text{J}2.0×10−4 J
  4. 2.0×102 W2.0\times10^{2}\ \text{W}2.0×102 W
Explanation: This question tests calculating average power from work per cycle and frequency, using P = W × f. Power equals work per unit time, and for cyclic processes, this becomes work per cycle times cycles per second. The actuator does 4.0×10⁻³ J per cycle at 50 Hz, so P = (4.0×10⁻³ J)(50 s⁻¹) = 2.0×10⁻¹ W = 0.20 W. Choice B appears to divide instead of multiply, while choice C gives units of energy (J) instead of power (W). For cyclic systems, multiply work per cycle by frequency to get average power, and always verify the result has units of watts.

Question 11

A physical therapist compares two ways of raising a patient’s leg of effective mass mmm by the same vertical height hhh: (1) slowly over 4 s and (2) quickly over 1 s. Assume the leg starts and ends at rest in both cases, and take g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2. The core principle tested is the distinction between work and power (W=mghW = mghW=mgh, Pavg=W/ΔtP_{\text{avg}} = W/\Delta tPavg​=W/Δt). Which statement is most consistent with these principles?

  1. Both methods require the same work mghmghmgh, but the quicker lift requires greater average power.
  2. The quicker lift requires more work because power increases with speed.
  3. Both methods require the same average power because the height change is the same.
  4. The slower lift requires greater average power because the force acts for a longer time.
Explanation: This question tests the distinction between work and power, where work depends on force and displacement while power depends on how quickly work is done. Work is a measure of energy transfer (W = mgh for vertical lifting), while power is the rate of energy transfer (P = W/Δt). Both methods require the same work W = mgh since the height change is identical, but the quicker lift requires greater average power: P₁ = mgh/1s versus P₂ = mgh/4s. Choice B incorrectly claims work depends on speed, while choice D reverses the power relationship. When comparing lifting scenarios, remember that work depends only on the height change, but power inversely depends on time taken.

Question 12

A researcher measures mechanical output during a short burst on a cycle ergometer. A subject produces 300 J300\ \text{J}300 J of mechanical work over 2.0 s2.0\ \text{s}2.0 s while pedaling. Using the definition of power, which prediction is most consistent with the measurement?

  1. Average power is 150 W150\ \text{W}150 W for the interval.
  2. Average power is 600 W600\ \text{W}600 W because power equals work times time.
  3. Average power is 0 W0\ \text{W}0 W because the work is positive only if speed increases.
  4. Average power is 300 N300\ \text{N}300 N because joules convert directly to newtons.
Explanation: This question tests the definition of power as work divided by time. Power is the rate at which work is done, calculated as P = W/t. Given 300 J of work done in 2.0 s, the average power is 300 J / 2.0 s = 150 W. Choice B incorrectly multiplies work by time instead of dividing, while Choice D confuses units (watts are not newtons). Choice C misunderstands that work can be done even at constant speed. When calculating power, always divide total work by total time, and remember that the unit of power is the watt (J/s).

Question 13

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 relationship W=FdW=FdW=Fd (force parallel to displacement). Which outcome is most consistent for the work done by the motor?

  1. The work doubles.
  2. The work is unchanged because speed is constant.
  3. The work is halved because the same force is spread over more distance.
  4. The work becomes negative because displacement increases.
Explanation: 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.

Question 14

A 5.0 kg5.0\ \text{kg}5.0 kg box is lifted vertically at constant speed by a rope with tension approximately equal to its weight. The box rises 1.0 m1.0\ \text{m}1.0 m. 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=9.8 m/s2g=9.8\ \text{m/s}^2g=9.8 m/s2.)

  1. Gravity does negative work of magnitude mghmghmgh.
  2. Gravity does positive work of magnitude mghmghmgh because the box moves upward.
  3. Gravity does zero work because the speed is constant.
  4. Gravity’s work depends on the rope tension, not on displacement.
Explanation: 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.

Question 15

During a vertical jump, a force plate measures that the net upward work done on a 60 kg60\ \text{kg}60 kg athlete’s center of mass during push-off is 240 J240\ \text{J}240 J. Apply the work–energy theorem to the center of mass. Which outcome is most consistent for the change in kinetic energy during push-off?

  1. The kinetic energy increases by 240 J240\ \text{J}240 J.
  2. The kinetic energy decreases by 240 J240\ \text{J}240 J because the athlete moves upward.
  3. The kinetic energy change must be zero because the athlete is still in contact with the ground.
  4. The kinetic energy change equals 240 W240\ \text{W}240 W because work converts to power.
Explanation: This question tests direct application of the work-energy theorem. The work-energy theorem states that net work equals change in kinetic energy: W_net = ΔKE. Given that net upward work is 240 J, the kinetic energy increases by exactly 240 J. Choice B incorrectly assumes upward motion means kinetic energy decreases. Choice C incorrectly assumes ground contact prevents kinetic energy change. Choice D confuses units of energy (joules) with power (watts). The work-energy theorem directly relates net work to kinetic energy change, regardless of direction of motion.

Question 16

A person carries a 10 kg10\ \text{kg}10 kg backpack horizontally at constant speed across a level hallway for 20 m20\ \text{m}20 m. The force the person exerts on the backpack is approximately vertical (supporting its weight). Using the definition of mechanical work, which statement is most consistent about the work the person does on the backpack during the horizontal walk?

  1. Approximately zero work, because the applied force is perpendicular to displacement.
  2. Positive work equal to mghmghmgh because the backpack has weight.
  3. Negative work because the backpack resists motion.
  4. Work equals mgdmgdmgd because distance traveled determines work regardless of direction.
Explanation: This question tests understanding that work requires force parallel to displacement. The person exerts an upward force to support the backpack's weight, but the displacement is horizontal. Since force and displacement are perpendicular (90° angle), the work done is W = Fd cos(90°) = 0. Choice B incorrectly calculates work using vertical force and horizontal distance. Choice D ignores the direction relationship between force and displacement. To calculate work correctly, only the component of force parallel to displacement contributes, and perpendicular forces do zero work.

Question 17

A microfluidic pump lifts blood (density ρ=1060 kg/m3\rho = 1060\ \text{kg/m}^3ρ=1060 kg/m3) through a vertical height difference of 0.50 m0.50\ \text{m}0.50 m at a volumetric flow rate of 1.0 mL/s1.0\ \text{mL/s}1.0 mL/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=9.8 m/s2g=9.8\ \text{m/s}^2g=9.8 m/s2; 1 mL=1×10−6 m31\ \text{mL}=1\times 10^{-6}\ \text{m}^31 mL=1×10−6 m3.)

  1. Power is proportional to ρghV˙\rho g h \dot{V}ρghV˙, so doubling V˙\dot{V}V˙ doubles required power
  2. Power is proportional to ρgV˙/h\rho g \dot{V}/hρgV˙/h, so doubling hhh halves required power
  3. Power is proportional to ρgh/V˙\rho g h/\dot{V}ρgh/V˙, so increasing flow rate decreases required power
  4. Power is independent of ρ\rhoρ because gravity acts equally on all fluids
Explanation: 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.

Question 18

A lab cart carrying a tissue sample is pulled along a horizontal track at constant speed by a motor applying a constant force of 5 N5\ \text{N}5 N over 4 m4\ \text{m}4 m. Principle tested: work by a constant force. Which statement is most consistent with the work done by the motor on the cart over this displacement?

  1. Work is 20 J20\ \text{J}20 J because W=FdW = FdW=Fd when force is parallel to displacement
  2. Work is 1.25 J1.25\ \text{J}1.25 J because work equals force divided by distance
  3. Work is 0 J0\ \text{J}0 J because speed is constant
  4. Work is 9 J9\ \text{J}9 J because work equals impulse, FtFtFt
Explanation: The skill being tested is calculating work done by a constant force. The principle is that work W = F d cosθ, where θ is the angle between force and displacement; for parallel force and constant speed, work equals force times distance. In this lab cart scenario, the motor's constant force is parallel to the horizontal displacement, so work is simply F d. Choice A is consistent because W = 5 N * 4 m = 20 J, matching the definition for parallel force. Choice C is incorrect because it claims zero work at constant speed, misunderstanding that work can be done without acceleration if opposing forces are present. In similar constant-force problems, multiply force component parallel to displacement by distance. Check if speed is constant to confirm no net work changes KE, but individual forces can still do work.

Question 19

A prosthetic actuator lifts a 2.0 kg2.0\ \text{kg}2.0 kg load vertically at constant speed through 0.40 m0.40\ \text{m}0.40 m in 2.0 s2.0\ \text{s}2.0 s. Principle tested: power as rate of doing work. Which prediction is most consistent with the average mechanical power delivered to the load? (Use g=9.8 m/s2g=9.8\ \text{m/s}^2g=9.8 m/s2.)

  1. ≈3.9 W\approx 3.9\ \text{W}≈3.9 W because power equals force times time
  2. ≈7.8 W\approx 7.8\ \text{W}≈7.8 W because P=mghtP = \frac{mgh}{t}P=tmgh​
  3. ≈19.6 W\approx 19.6\ \text{W}≈19.6 W because P=mgtP = mg tP=mgt
  4. ≈0.8 W\approx 0.8\ \text{W}≈0.8 W because constant speed implies zero work
Explanation: The skill being tested is calculating average power in lifting against gravity. The principle is that average power P = W/t, where work W = mgh for vertical lifts at constant speed with no KE change. In this prosthetic actuator, the load is lifted at constant speed, so power is mgh/t. Choice B is consistent because P ≈ (2)(9.8)(0.4)/2 ≈ 7.8 W. Choice D is incorrect because it claims zero power at constant speed, misunderstanding that work is done against gravity despite no acceleration. In similar lifting problems, calculate work as mgh and divide by time. Confirm constant speed to neglect KE changes.

Question 20

A patient pushes a walker with a horizontal force of 30 N30\ \text{N}30 N over 5 m5\ \text{m}5 m on a level floor. The force is applied at a 60∘60^\circ60∘ angle above the horizontal (handle angled upward). Principle tested: work by a force at an angle. Which expression is most consistent with the mechanical work done by the patient on the walker (ignoring vertical displacement)?

  1. W=Fdsin⁡60∘W = Fd\sin 60^\circW=Fdsin60∘
  2. W=Fdcos⁡60∘W = Fd\cos 60^\circW=Fdcos60∘
  3. W=F/dW = F/dW=F/d
  4. W=Fdcos⁡30∘W = Fd\cos 30^\circW=Fdcos30∘ because the vertical component does the work
Explanation: The skill being tested is calculating work when force is at an angle to displacement. The principle is that work W = F d cosθ, using the component of force parallel to displacement. In this walker scenario, the force at 60° above horizontal means the parallel component is F cos60°, so W = F d cos60°. Choice B is consistent because it uses cos60° for the horizontal component doing work on level floor. Choice A is incorrect because it uses sin60°, misunderstanding that the vertical component does no work without vertical displacement. For similar angled-force problems, project force onto displacement direction. Verify by noting θ is between force and displacement vectors.