Conservation of Energy and Mechanical Advantage (4A)
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MCAT Chemical and Physical Foundations of Biological Systems › Conservation of Energy and Mechanical Advantage (4A)
A clinician uses a non-ideal block-and-tackle to lift a $15\ \text{kg}$ rehabilitation weight stack by $0.40\ \text{m}$. The rope is pulled $1.6\ \text{m}$ with an average force of $50\ \text{N}$. Which statement is most consistent with energy conservation and the distinction between mechanical advantage and efficiency? (Use $g=10\ \text{m/s}^2$.)
Because the rope is pulled a longer distance than the weight rises, the system must have efficiency greater than 100%.
Input work is $80\ \text{J}$ while the weight gains $60\ \text{J}$ of potential energy; the efficiency is $60/80=75%$.
Input work equals output work in any pulley, so the weight must gain $80\ \text{J}$ of potential energy regardless of mass.
Mechanical advantage equals efficiency, so an efficiency of 75% implies a mechanical advantage of 0.75.
Explanation
This question tests understanding of the distinction between mechanical advantage and efficiency in real machines. Input work = 50 N × 1.6 m = 80 J. The weight gains potential energy = mgh = 15 × 10 × 0.40 = 60 J. Efficiency = (useful output work)/(input work) = 60/80 = 75%. The mechanical advantage is the ratio of distances: 1.6/0.40 = 4, which is independent of efficiency. Choice B is correct because it accurately calculates both input work and output work, determining efficiency as their ratio. Choice D incorrectly equates mechanical advantage with efficiency, which are distinct concepts. When analyzing real machines, remember that mechanical advantage relates to force/distance ratios, while efficiency measures energy conservation.
A $0.50\ \text{kg}$ instrument is lowered at constant speed by a motorized winch through $2.0\ \text{m}$. The motor acts as a brake and delivers $6\ \text{J}$ of electrical energy to a resistor during the descent. Which prediction best aligns with conservation of energy? (Use $g=10\ \text{m/s}^2$.)
The gravitational potential energy decreases by $1\ \text{J}$ because only energy converted to electricity counts as lost potential energy.
The gravitational potential energy decrease is $20\ \text{J}$ because energy depends on distance traveled, not vertical displacement.
The gravitational potential energy decreases by $10\ \text{J}$; $6\ \text{J}$ is converted to electrical energy and the remaining $4\ \text{J}$ is dissipated as heat (e.g., friction).
The gravitational potential energy decreases by $6\ \text{J}$ and the remaining $4\ \text{J}$ appears as increased kinetic energy at the bottom.
Explanation
This question tests understanding of energy conservation when multiple energy transformations occur. The instrument loses gravitational potential energy = mgh = 0.50 × 10 × 2.0 = 10 J. Since the speed is constant, kinetic energy doesn't change. Of the 10 J lost, 6 J is converted to electrical energy by the motor acting as a generator. The remaining 4 J must be dissipated as heat due to friction or other losses. Choice A is correct because it accounts for all energy transformations: 10 J of gravitational potential energy converts to 6 J electrical plus 4 J heat. Choice C incorrectly suggests kinetic energy increases despite constant speed. When analyzing energy conversions, ensure all input energy is accounted for in various output forms.
A student designs a hand-operated lever to open a tight vial cap. In prototype X, the effort arm is doubled while the load arm is unchanged. The cap must rotate through the same small angle to open in both designs, and losses are negligible. Which prediction would align with conservation of energy and mechanical advantage?
Prototype X requires the same effort force but half the effort distance, so the input work decreases.
Prototype X requires greater effort force because increased mechanical advantage always increases required input force.
Prototype X requires half the effort force and the same effort distance, so the lever produces extra energy.
Prototype X requires half the effort force but twice the effort distance, so the input work is approximately unchanged.
Explanation
This question tests understanding of how mechanical advantage affects the force-distance trade-off while conserving energy. Doubling the effort arm while keeping the load arm constant doubles the mechanical advantage. This means the required effort force is halved, but the effort must move through twice the distance to rotate the cap the same angle. The work input (force × distance) remains approximately the same: (F/2) × (2d) = Fd. Choice B is correct because it recognizes that mechanical advantage trades force for distance while conserving total work done. Choice C violates energy conservation by suggesting the same distance with less force. When analyzing lever modifications, remember that mechanical advantage cannot create energy—it only redistributes the force-distance relationship.
In a biomechanics lab, a researcher uses an ideal (massless, frictionless) 2:1 pulley system to lift a $20\ \text{kg}$ instrument tray vertically by $0.50\ \text{m}$ at constant speed. The free end of the rope is pulled downward. Which prediction would align with conservation of energy and mechanical advantage for this system? (Use $g=10\ \text{m/s}^2$.)
Pulling $100\ \text{N}$ through $0.50\ \text{m}$ raises the tray $0.25\ \text{m}$, delivering $50\ \text{J}$ of work to the tray.
Pulling $200\ \text{N}$ through $0.50\ \text{m}$ raises the tray $0.50\ \text{m}$, delivering $100\ \text{J}$ of work to the tray.
Pulling $100\ \text{N}$ through $1.0\ \text{m}$ raises the tray $0.50\ \text{m}$, delivering $100\ \text{J}$ of work to the tray.
Pulling $50\ \text{N}$ through $0.50\ \text{m}$ raises the tray $0.50\ \text{m}$, delivering $25\ \text{J}$ of work to the tray.
Explanation
This question tests understanding of conservation of energy and mechanical advantage in pulley systems. In an ideal 2:1 pulley system, the mechanical advantage means you pull with half the force but through twice the distance compared to direct lifting. To lift a 20 kg mass (weight = 200 N) by 0.50 m requires 100 J of work (W = mgh = 20 × 10 × 0.50). With a 2:1 pulley, you must pull with 100 N force through 1.0 m, giving input work of 100 J, which equals the output work. Choice A is correct because it shows energy conservation: 100 N × 1.0 m = 100 J input equals the 100 J needed to raise the tray. Choice B incorrectly uses 200 N force, which would be needed without mechanical advantage. When analyzing pulley systems, verify that input work equals output work for ideal systems, and remember that mechanical advantage trades force for distance.
A physical therapist models elbow flexion as a lever. A $50\ \text{N}$ load in the patient’s hand acts $0.30\ \text{m}$ from the elbow joint. The biceps inserts $0.040\ \text{m}$ from the elbow and pulls approximately perpendicular to the forearm. Neglect forearm weight. Which statement best illustrates the mechanical advantage of this lever arrangement?
The biceps must do less work than the load rises because the lever provides mechanical advantage greater than 1 in humans.
Because the biceps attaches closer to the joint than the load, the biceps force must exceed $50\ \text{N}$ to hold the load static.
Mechanical advantage depends only on muscle efficiency, so changing insertion distance does not affect required biceps force.
Because the biceps attaches closer to the joint than the load, the biceps force can be less than $50\ \text{N}$ while holding the load static.
Explanation
This question tests understanding of mechanical advantage in lever systems, specifically the human forearm as a third-class lever. In a lever system, torque balance requires that force × distance from fulcrum must be equal on both sides. The load creates a torque of 50 N × 0.30 m = 15 N·m about the elbow joint. To balance this, the biceps must create an equal torque: F_biceps × 0.040 m = 15 N·m, giving F_biceps = 375 N. Choice A is correct because the biceps attaches much closer to the joint (0.040 m) than the load (0.30 m), requiring a force much greater than 50 N to maintain equilibrium. Choice B incorrectly suggests the biceps force could be less than the load, violating torque balance. When analyzing biological levers, remember that muscles often attach close to joints, requiring large forces but allowing greater range of motion.
A researcher tests a pulley used to raise an aquarium water reservoir for a zebrafish facility. The reservoir mass is $10\ \text{kg}$. The technician pulls the rope with a constant force of $60\ \text{N}$ through $2.0\ \text{m}$, raising the reservoir by $1.0\ \text{m}$. Which statement is most consistent with conservation of energy for this non-ideal system? (Use $g=10\ \text{m/s}^2$.)
The work input is $60\ \text{J}$ and the gravitational potential energy gained is $100\ \text{J}$; the pulley amplifies energy by mechanical advantage.
The work input is $120\ \text{J}$ and the gravitational potential energy gained is $100\ \text{J}$; the $20\ \text{J}$ difference is dissipated (e.g., friction/heat).
Because mechanical advantage is 2, the gravitational potential energy gained must be $240\ \text{J}$ for a $120\ \text{J}$ input.
The reservoir gains $60\ \text{J}$ of gravitational potential energy; the remaining input energy is stored as elastic potential in the rope.
Explanation
This question tests understanding of energy conservation in non-ideal mechanical systems where friction or other losses occur. The work input is force × distance = 60 N × 2.0 m = 120 J. The gravitational potential energy gained by the reservoir is mgh = 10 × 10 × 1.0 = 100 J. Since input work (120 J) exceeds output work (100 J), the difference of 20 J must be dissipated as heat, sound, or other non-recoverable forms. Choice B is correct because it accurately accounts for both the input work and output work, identifying the 20 J loss to friction or other dissipative forces. Choice C incorrectly suggests the pulley can amplify energy, violating conservation laws. When analyzing real systems, always expect input work to exceed useful output work due to inevitable losses.
During a vertical jump, a $70\ \text{kg}$ athlete’s center of mass rises by $0.50\ \text{m}$. Ignoring air resistance, which statement is most consistent with conservation of energy for the upward motion after takeoff? (Use $g=10\ \text{m/s}^2$.)
Gravitational potential energy increases by $700\ \text{J}$ because energy depends on velocity as well as height.
Gravitational potential energy increases by $350\ \text{J}$, so kinetic energy must also increase by $350\ \text{J}$ during ascent.
Gravitational potential energy does not change after takeoff because no external work is done on the athlete.
Gravitational potential energy increases by $350\ \text{J}$, implying the athlete’s kinetic energy decreases by $350\ \text{J}$ during ascent.
Explanation
This question tests understanding of energy conservation during projectile motion after takeoff. During the upward phase of a jump, the athlete's total mechanical energy remains constant (ignoring air resistance). As the center of mass rises 0.50 m, gravitational potential energy increases by mgh = 70 × 10 × 0.50 = 350 J. Since total mechanical energy is conserved and potential energy increases, kinetic energy must decrease by exactly 350 J. Choice B is correct because it recognizes that the increase in potential energy comes from a corresponding decrease in kinetic energy, maintaining constant total energy. Choice A incorrectly suggests both energies increase, violating conservation. When analyzing vertical motion, remember that kinetic and potential energy trade off while their sum remains constant.
A clinician uses a forearm crutch to raise a patient’s body slightly during ambulation. Modeling the crutch as an ideal lever, the hand applies a downward force $F_h$ at a point $30\ \text{cm}$ from the crutch tip (pivot on the ground), while the patient’s weight supported by the crutch acts $10\ \text{cm}$ from the tip. Which prediction aligns with torque balance and mechanical advantage?
$F_h$ is about one-third of the supported weight, because the hand is farther from the pivot
$F_h$ is about three times the supported weight, because the hand is farther from the pivot
$F_h$ can be zero if the crutch is rigid, because rigidity replaces force
$F_h$ equals the supported weight, because levers only change direction of force
Explanation
This question tests understanding of conservation of energy and mechanical advantage in physical systems. Conservation of energy states that energy cannot be created or destroyed, only transformed. In the given system, the crutch functions as a lever with torque balanced around the pivot. Choice A is correct because the longer moment arm at the hand (30 cm vs. 10 cm) makes F_h one-third the supported weight. Choice B is incorrect because it reverses the moment arm effect, predicting a larger force. When analyzing similar systems, apply torque equilibrium and note mechanical advantage reduces effort force with longer effort arms.
A lab measures the work required to lift a $150\ \text{N}$ tissue-sample container by $0.50\ \text{m}$ using an inclined plane of length $2.0\ \text{m}$. Friction is non-negligible and the measured input work is $100\ \text{J}$. Which statement is most consistent with conservation of energy?
The gravitational potential energy gain is $100\ \text{J}$, so friction must be zero
The gravitational potential energy gain is $75\ \text{J}$, so about $25\ \text{J}$ is dissipated as heat
The gravitational potential energy gain is $300\ \text{J}$, so the system created energy
The gravitational potential energy gain is $50\ \text{J}$, because longer ramps reduce energy needed
Explanation
This question tests understanding of conservation of energy and mechanical advantage in physical systems. Conservation of energy states that energy cannot be created or destroyed, only transformed. In the given system, input work exceeds potential energy gain due to friction on the ramp. Choice A is correct because PE gain is 150×0.5=75 J, with 100-75=25 J dissipated. Choice B is incorrect because it suggests energy creation, violating conservation. When analyzing similar systems, compare input and output work to find efficiency, and account for dissipative losses.
A physical therapy device uses a spring-loaded platform to assist a patient stepping up. The platform compresses a spring (spring constant $k=800\ \text{N/m}$) by $0.10\ \text{m}$ and then releases, lifting a $20\ \text{kg}$ load vertically with negligible losses. Which maximum lift height is most consistent with energy conservation? (Use $g=9.8\ \text{m/s}^2$.)
$0.02\ \text{m}$, because $\tfrac12 kx^2 \approx mgh$
$0.10\ \text{m}$, because the spring compression equals the lift height
$0.20\ \text{m}$, because $kx\approx mgh$
$2.0\ \text{m}$, because springs can amplify energy through mechanical advantage
Explanation
This question tests understanding of conservation of energy and mechanical advantage in physical systems. Conservation of energy states that energy cannot be created or destroyed, only transformed. In the given system, spring potential energy converts to gravitational potential for the load. Choice A is correct because 1/2 $kx^2$ ≈ mgh yields h≈0.02 m from the values. Choice B is incorrect because it uses kx ≈ mgh, omitting the 1/2 and overestimating height. When analyzing similar systems, balance elastic and gravitational energies, and note no mechanical advantage amplification beyond conservation.