Conservation of Energy and Mechanical Advantage (4A)
Help Questions
MCAT Chemical and Physical Foundations of Biological Systems › Conservation of Energy and Mechanical Advantage (4A)
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 $10\ \text{J}$; $6\ \text{J}$ is converted to electrical energy and the remaining $4\ \text{J}$ is dissipated as heat (e.g., friction).
The gravitational potential energy decrease is $20\ \text{J}$ because energy depends on distance traveled, not vertical displacement.
The gravitational potential energy decreases by $1\ \text{J}$ because only energy converted to electricity counts as lost potential energy.
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.
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 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.
Gravitational potential energy increases by $350\ \text{J}$, so kinetic energy must also increase 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?
$F_h$ equals the supported weight, because levers only change direction of force
$F_h$ can be zero if the crutch is rigid, because rigidity replaces force
$F_h$ is about three times the supported weight, because the hand is farther from the pivot
$F_h$ is about one-third of the supported weight, because the hand is farther from the pivot
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 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 $g=9.8\ \text{m/s}^2$.)
$2.0\ \text{m}$, because springs can amplify energy through mechanical advantage
$0.10\ \text{m}$, because the spring compression equals the lift height
$0.20\ \text{m}$, because $kx\approx mgh$
$0.02\ \text{m}$, because $\tfrac12 kx^2 \approx mgh$
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.
A lab uses a single fixed pulley to redirect a force when lifting a $80\ \text{N}$ container at constant speed. The pulley axle has friction such that efficiency is $90%$. Which applied force is most consistent with energy conservation?
$72\ \text{N}$, because $90%$ efficiency reduces the needed force
$80\ \text{N}$, because fixed pulleys always require exactly the load force even with friction
$89\ \text{N}$, because $F\approx \dfrac{80}{0.90}$ for a fixed pulley with losses
$160\ \text{N}$, because a pulley doubles the distance so it must double the 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, fixed pulley redirects force with efficiency loss. Choice C is correct because F≈80/0.90=89 N accounts for friction reducing output. Choice A is incorrect because efficiency reduces, not increases, effective MA. When analyzing similar systems, use eff=workout/workin to find actual force, noting fixed pulleys have MA=1 ideally.
A $0.50\ \text{kg}$ sample holder is launched straight upward by a pneumatic actuator and rises to a maximum height of $2.0\ \text{m}$ above the launch point. Neglecting air resistance, which initial kinetic energy is most consistent with conservation of energy? (Use $g=9.8\ \text{m/s}^2$.)
$19.6\ \text{J}$, because $K_i=2mgh$ for upward motion
$4.9\ \text{J}$, because $K_i=\tfrac12 mgh$
$0.25\ \text{J}$, because kinetic energy decreases with height
$9.8\ \text{J}$, because $K_i=mgh$
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, initial kinetic energy converts to potential at maximum height. Choice A is correct because Ki=mgh=9.8 J matches the energy transformation. Choice B is incorrect because it halves the value unnecessarily. When analyzing similar systems, set initial KE equal to max PE, ignoring air resistance for conservation.
A $1.0\ \text{kg}$ sensor package slides down a frictionless track from a height of $0.80\ \text{m}$ and then enters a horizontal section. Which prediction is most consistent with conservation of energy regarding the speed on the horizontal section (immediately after the descent)? (Use $g=9.8\ \text{m/s}^2$.)
It is constant and nonzero because the kinetic energy at the bottom is conserved on the horizontal section
It decreases because conservation of energy requires kinetic energy to decay with time
It is zero because all energy becomes potential at the bottom
It increases because potential energy continues to decrease on a horizontal surface
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, potential energy becomes kinetic on descent, conserved on horizontal. Choice B is correct because KE remains constant on frictionless horizontal after gain. Choice A is incorrect because KE is maximum at bottom. When analyzing similar systems, track energy forms across sections, noting no PE change horizontally.
An ergometer flywheel (moment of inertia $I$) is spun up and then allowed to lift a small mass $m$ via a string wrapped around the axle (radius $r$). Neglecting friction and string slip, which statement is consistent with conservation of energy as the mass rises?
The flywheel’s rotational kinetic energy increases because lifting requires added kinetic energy
The flywheel’s rotational kinetic energy decreases as the mass gains gravitational potential energy
The mass can rise without slowing the flywheel because tension is an internal force
Energy is not conserved because rotational motion cannot be converted to potential energy
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, flywheel rotational KE transfers to mass PE. Choice A is correct because KE decrease provides the PE gain. Choice B is incorrect because lifting reduces source KE. When analyzing similar systems, track energy transfer from rotational to potential, ensuring no creation or loss.
A researcher uses a wedge to slightly separate two tightly fitted components in a lab instrument. The wedge is pushed horizontally with force $F$ over distance $d$, raising one component vertically by a smaller distance $h$ with output force $F_{out}$. Ignoring friction, which relationship is most consistent with energy conservation in simple machines?
$F,d > F_{out},h$ must always hold even without friction, because machines create extra work
$F,d \approx F_{out},h$, so larger $F_{out}$ corresponds to smaller $h$ for fixed input work
$F,h \approx F_{out},d$, so larger $F_{out}$ corresponds to larger $h$ for fixed input work
$F_{out}$ is independent of displacement because wedges only change force direction
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 wedge system, the input work done by pushing the wedge horizontally with force F over distance d equals the output work done in raising the component vertically by height h with force F_out, assuming no friction, so F d = F_out h. Choice A is correct because it reflects the principle of energy conservation by showing that input work approximates output work, and for fixed input work, a larger output force F_out corresponds to a smaller output displacement h. Choice B is incorrect because it swaps the displacements, incorrectly suggesting that larger F_out corresponds to larger h, which violates the inverse relationship in mechanical advantage. When analyzing similar systems, ensure that work input equals work output in ideal machines, and identify that mechanical advantage provides greater force but over reduced distance without generating extra energy.
In a biomechanics study, a subject performs a vertical jump. A motion-capture system estimates the subject’s center of mass (mass $m=70\ \text{kg}$) rises by $0.45\ \text{m}$ from takeoff to peak height, and air resistance during ascent is negligible. Using conservation of mechanical energy, which prediction is most consistent with the subject’s speed at takeoff? (Use $g=9.8\ \text{m/s}^2$.)
About $3.0\ \text{m/s}$, because $\tfrac12 mv^2 \approx mgh$
About $6.6\ \text{m/s}$, because $mv^2 \approx mgh$
About $9.8\ \text{m/s}$, because the takeoff speed must equal $g$ for a vertical rise
About $0.45\ \text{m/s}$, because speed must match height numerically when $g\approx 10$
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, kinetic energy at takeoff converts to gravitational potential energy at peak height during the jump. Choice A is correct because it accurately uses 1/2 $mv^2$ ≈ mgh to calculate v ≈ 3.0 m/s from the given height and g. Choice B is incorrect because it uses $mv^2$ ≈ mgh, omitting the 1/2 and overestimating speed. When analyzing similar systems, ensure energy input equals energy output across transformations, and note mechanical advantage may apply in leveraged motions but not purely in free jumps.