Work, Energy, and Power (4A)
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MCAT Chemical and Physical Foundations of Biological Systems › Work, Energy, and Power (4A)
In a cycling ergometer test, two subjects each perform $300\ \text{J}$ of mechanical work on the pedals. Subject 1 completes the work in $10\ \text{s}$; Subject 2 completes the same work in $20\ \text{s}$. Using the definition of power $P = W/t$, which prediction is most consistent with the data?
Subject 1 has lower power output because power depends only on force, not on time.
Both subjects have the same power output because they perform the same total work.
Subject 2 has higher power output because longer time allows more energy to be delivered per second.
Subject 1 has higher power output because the same work is completed in less time.
Explanation
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.
During a physical therapy session, a patient lifts a $4.0\ \text{kg}$ dumbbell vertically upward by $0.50\ \text{m}$ at constant speed. Neglecting air resistance and taking $g = 9.8\ \text{m/s}^2$, apply the work–energy theorem to the dumbbell. Which statement is most consistent with the net work on the dumbbell?
The net work on the dumbbell equals $mgh$ because net work always equals the change in potential energy.
The net work on the dumbbell is zero because its kinetic energy does not change at constant speed.
The net work on the dumbbell is $mg$ because work and force are equivalent for vertical motion.
The net work on the dumbbell is positive because the dumbbell is moving upward.
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.
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 $2\ \text{N}$ over $0.10\ \text{m}$ in $1.0\ \text{s}$. Protocol B applies $1\ \text{N}$ over $0.20\ \text{m}$ in $2.0\ \text{s}$. Using work $W = Fd$ (force parallel to displacement) and power $P = W/t$, which prediction is most consistent?
Both protocols deliver the same work, but Protocol A has higher average power.
Protocol B delivers greater average power because it lasts longer.
Protocol A delivers greater work because it uses a larger force.
Both protocols deliver the same work and the same 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.
In an isolated muscle preparation, a weight is attached to a muscle tendon and lifted at constant speed through a vertical distance $h$. The investigator repeats the trial with the same weight but doubles the lift height to $2h$. Neglecting losses, apply conservation of energy to predict how the required mechanical work changes. Which outcome is most consistent?
The required work halves because the same force is applied over a longer distance.
The required work quadruples because potential energy depends on $h^2$.
The required work is unchanged because the weight and speed are unchanged.
The required work doubles because the increase in gravitational potential energy is proportional to height.
Explanation
This question tests conservation of energy to predict how work requirements change when lifting height is varied. Conservation of energy requires that the work done by the muscle equals the change in gravitational potential energy when lifting at constant speed with no losses. When lifting a weight through height h, the required work is W = mgh, and when lifting through height 2h, the required work becomes W = mg(2h) = 2mgh. The correct answer recognizes that doubling the height doubles the required work because gravitational potential energy is directly proportional to height. Choice D incorrectly suggests potential energy depends on h², confusing the linear relationship ΔPE = mgΔh with kinematic equations for free fall. To predict work requirements for vertical lifting, remember that work equals the change in gravitational potential energy: W = mgΔh, where work is directly proportional to the height change.
A biomechanical model treats a sprinter’s center of mass as a particle. Over a short interval, the net external work done on the sprinter is measured as $-50\ \text{J}$ (e.g., due to braking forces). By the work–energy theorem, which change in kinetic energy is most consistent?
The kinetic energy decreases by $50\ \text{N}$ because work is measured in newtons during running.
The kinetic energy decreases by $50\ \text{J}$ because $W_{\text{net}} = \Delta K$.
The kinetic energy increases by $50\ \text{J}$ because negative work implies acceleration forward.
The kinetic energy change is zero because work changes potential energy, not kinetic energy.
Explanation
This question tests the work-energy theorem with negative net work, requiring understanding of how work's sign affects kinetic energy changes. The work-energy theorem (W_net = ΔK) applies regardless of whether net work is positive or negative: positive work increases kinetic energy while negative work decreases it. With W_net = -50 J, the change in kinetic energy must be ΔK = -50 J, meaning kinetic energy decreases by 50 J. The correct answer properly applies the work-energy theorem to conclude that kinetic energy decreases by 50 J when net work is -50 J. Choice A incorrectly interprets negative work as causing acceleration forward, when negative work actually removes kinetic energy and causes deceleration. When applying the work-energy theorem, always match the sign: positive net work increases kinetic energy, negative net work decreases it, with the numerical values being equal.
A handheld ultrasound probe is pressed against skin with a force of $6\ \text{N}$ while sliding $0.30\ \text{m}$ along the surface. The applied force is perpendicular to the direction of motion (idealized). Using the definition of work $W = Fd\cos\theta$, which outcome is most consistent for the work done by the applied force on the probe during the slide?
The work is $6\ \text{J}$ because work has the same units as force.
The work is $0\ \text{J}$ because the force is perpendicular to the displacement.
The work is negative because perpendicular forces always remove energy from a system.
The work is $1.8\ \text{J}$ because work equals force times distance for any angle.
Explanation
This question tests the definition of work when force and displacement are not aligned, requiring application of W = Fd cos θ. Work is defined as the dot product of force and displacement vectors, which equals Fd cos θ where θ is the angle between them. When force is perpendicular to displacement (θ = 90°), cos 90° = 0, making the work W = Fd cos 90° = 0 regardless of the magnitudes of force and displacement. The correct answer recognizes that perpendicular force does zero work because cos 90° = 0 in the work formula. Choice B incorrectly ignores the angle dependence, treating work as simply force times distance, which only applies when force and displacement are parallel. To calculate work correctly, always identify the angle between force and displacement vectors: parallel (θ = 0°) gives maximum work, antiparallel (θ = 180°) gives negative work, and perpendicular (θ = 90°) gives zero work.
A stair-climb test estimates mechanical power by timing how quickly a subject raises their center of mass by a vertical height $h$ against gravity. Two subjects have the same mass $m$ and climb the same height $h$, but Subject X completes the climb in half the time of Subject Y. Neglecting losses and using $P = W/t$ with $W \approx mgh$, which prediction is most consistent with the power relationship?
Subject X produces twice the average power because the same work is done in half the time.
Subject X produces the same average power because both subjects do the same work.
Subject X produces half the average power because completing the climb faster reduces total work.
Subject X produces twice the average work because power depends only on time, not on height.
Explanation
This question tests the relationship between power, work, and time when comparing subjects performing the same task at different rates. Power is the rate of doing work (P = W/t), and when the same work is done in different times, power is inversely proportional to time. Both subjects do the same work W = mgh (same mass, same height), but Subject X completes it in time t/2 while Subject Y takes time t, giving P_X = mgh/(t/2) = 2mgh/t = 2P_Y. The correct answer recognizes that halving the time doubles the power when work is constant, reflecting the inverse relationship between power and time. Choice A incorrectly suggests that faster completion reduces work, confusing the constancy of work (determined by mgh) with the variability of power (determined by completion time). When comparing power outputs for the same task, remember that P = W/t means faster completion (smaller t) yields higher power for fixed work.
In a treadmill study of gait rehabilitation, a 70-kg patient walks up a ramp of height $h = 1.5\ \text{m}$ 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 $g = 9.8\ \text{m/s}^2$). Which outcome is most consistent with the principle?
The patient’s gravitational potential energy is unchanged because constant speed implies zero work by muscles.
The patient’s gravitational potential energy increases by $mgh$ because chemical energy is converted into potential energy.
The patient’s gravitational potential energy decreases by $mgh$ because the ramp provides an upward normal force.
The patient’s gravitational potential energy increases by $mg/h$ 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.
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 $12\ \text{J}$ over the movement (losses neglected). By the work–energy theorem, which statement is most consistent with the measurement?
The net work done on the lower leg is $0\ \text{J}$ because internal muscle forces cannot do work on the body.
The net work done on the lower leg is $-12\ \text{J}$ because increasing kinetic energy requires negative work.
The net work done on the lower leg is $12\ \text{N}$ because work and force have the same units in biomechanics.
The net work done on the lower leg is $12\ \text{J}$ because $W_{\text{net}} = \Delta K$.
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.
A lab models blood flow using a pump that raises $V = 5.0\ \text{mL}$ of fluid (density $\rho = 1000\ \text{kg/m}^3$) by a vertical height $h = 0.80\ \text{m}$ at constant speed. Neglecting losses and using $g = 9.8\ \text{m/s}^2$, apply conservation of mechanical energy to identify the mechanical energy change of the fluid. Which statement is most consistent?
The fluid’s mechanical energy decreases by $mgh$ because the pump does negative work on the fluid.
The fluid’s mechanical energy is unchanged because constant speed implies zero change in potential energy.
The fluid’s mechanical energy increases by $\rho g/h$ because density replaces mass in potential energy.
The fluid’s mechanical energy increases by $mgh$ because gravitational potential energy increases.
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.