MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4a Energy Conservation Mechanical Advantage

Study 4a Energy Conservation Mechanical Advantage in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

MCAT Chemical and Physical Foundations of Biological Systems

4a Energy Conservation Mechanical Advantage

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What is the definition of mechanical energy EmechE_{mech} in terms of KK and UU?

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ANSWER

Emech=K+UE_{mech} = K + U. Mechanical energy sums kinetic and potential energies, representing the total energy convertible between motion and position without losses.

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Flashcard 1: What is the definition of mechanical energy EmechE_{mech} in terms of KK and UU?

Answer: Emech=K+UE_{mech} = K + U. Mechanical energy sums kinetic and potential energies, representing the total energy convertible between motion and position without losses.

Flashcard 2: Identify the input force needed to lift a load FoutF_{out} with an ideal machine having MA=4MA = 4.

Answer: Fin=Fout4F_{in} = \frac{F_{out}}{4}. For ideal machines, mechanical advantage inversely relates input and output forces, requiring one-fourth force to lift with MA=4.

Flashcard 3: State the formula for kinetic energy of a mass mm moving at speed vv.

Answer: K=12mv2K = \frac{1}{2}mv^2. This equation calculates the energy of motion, arising from integrating force over distance for constant acceleration.

Flashcard 4: Find the compression xx of a spring that stops a mass mm moving at speed vv on a frictionless surface.

Answer: x=vmkx = v\sqrt{\frac{m}{k}}. Equating initial kinetic energy to stored elastic potential gives the deformation where the spring constant balances mass and velocity.

Flashcard 5: Find the work done by kinetic friction over distance dd on a level surface with coefficient μk\mu_k.

Answer: Wf=μkmgdW_f = -\mu_k mgd. Friction work is negative as it opposes motion, dissipating energy proportional to coefficient, weight, and distance on a flat surface.

Flashcard 6: What is the sign of work done by kinetic friction on a sliding object moving forward?

Answer: Negative work. Kinetic friction opposes motion, so the angle between force and displacement is 180°, yielding a negative cosine and energy dissipation.

Flashcard 7: State the formula for power in terms of work WW done over time interval Δt\Delta t.

Answer: P=WΔtP = \frac{W}{\Delta t}. Power represents the rate of work done, averaging energy transfer over time for non-instantaneous processes.

Flashcard 8: State the magnitude of kinetic friction force in terms of μk\mu_k and normal force NN.

Answer: fk=μkNf_k = \mu_k N. Kinetic friction is proportional to the normal force, with μk\mu_k as the coefficient depending on surface properties.

Flashcard 9: Identify the ideal MA for an ideal movable pulley with one pulley supporting the load by two rope segments.

Answer: MA=2MA = 2. The movable pulley halves the required input force by distributing the load across two rope segments ideally.

Flashcard 10: State the efficiency formula in terms of actual mechanical advantage (AMA) and IMA.

Answer: η=AMAIMA\eta = \frac{AMA}{IMA}. This relates actual performance (AMA) to theoretical maximum (IMA), highlighting losses in real machines.

Flashcard 11: Find the speed at the bottom after dropping from rest through height hh with no friction.

Answer: v=2ghv = \sqrt{2gh}. Conservation of energy converts gravitational potential to kinetic, yielding speed independent of mass for free fall near Earth.

Flashcard 12: State the instantaneous power formula for a force F\vec{F} acting on a moving object with velocity v\vec{v}.

Answer: P=FvP = \vec{F}\cdot\vec{v}. Instantaneous power is the dot product capturing the rate of energy transfer at a specific moment for varying forces or velocities.

Flashcard 13: State the formula for work done by a constant force FF over displacement dd at angle θ\theta.

Answer: W=FdcosθW = Fd\cos\theta. Work measures energy transfer by force along displacement, with cosθ\cos\theta accounting for the component parallel to motion.

Flashcard 14: Identify the energy conservation equation for a system including nonconservative work WncW_{nc}.

Answer: Ki+Ui+Wnc=Kf+UfK_i + U_i + W_{nc} = K_f + U_f. This equation accounts for energy input or dissipation by nonconservative forces, balancing initial and final mechanical energies.

Flashcard 15: Identify the condition under which mechanical energy is conserved in a system.

Answer: Only conservative forces do work (no nonconservative work). Conservation holds when work is path-independent, as conservative forces like gravity store energy as potential without dissipation.

Flashcard 16: State the work–energy theorem relating net work WnetW_{net} to kinetic energy change.

Answer: Wnet=ΔKW_{net} = \Delta K. The theorem states that net work on an object equals its change in kinetic energy, derived from Newton's second law integrated over displacement.

Flashcard 17: State the efficiency formula for a machine in terms of output and input work.

Answer: η=WoutWin\eta = \frac{W_{out}}{W_{in}}. Efficiency quantifies the fraction of input work converted to useful output, less than 1 due to losses like friction.

Flashcard 18: Identify the ideal MA for an ideal fixed pulley (single pulley attached to the ceiling).

Answer: MA=1MA = 1. A fixed pulley changes force direction but not magnitude, resulting in no net force amplification ideally.

Flashcard 19: What is the definition of ideal mechanical advantage (IMA) using distances moved?

Answer: IMA=dindoutIMA = \frac{d_{in}}{d_{out}}. Ideal mechanical advantage assumes no energy losses, equating to the ratio of input to output distances based on geometry.

Flashcard 20: What is the ideal MA of an ideal block-and-tackle equal to, in terms of supporting rope segments nn?

Answer: MA=nMA = n. In a block-and-tackle system, mechanical advantage equals the number of load-supporting ropes, amplifying force proportionally.

Flashcard 21: State the relationship between nonconservative work WncW_{nc} and mechanical energy change ΔEmech\Delta E_{mech}.

Answer: Wnc=ΔEmechW_{nc} = \Delta E_{mech}. Nonconservative forces like friction change mechanical energy through path-dependent work, equaling the net gain or loss in EmechE_{mech}.

Flashcard 22: State the formula for gravitational potential energy near Earth for a mass at height hh.

Answer: Ug=mghU_g = mgh. This formula quantifies the energy stored due to an object's position in a gravitational field, derived from the work done against gravity over height hh.

Flashcard 23: What is the definition of mechanical advantage (MA) for a simple machine using forces?

Answer: MA=FoutFinMA = \frac{F_{out}}{F_{in}}. Mechanical advantage measures force amplification, defined as the ratio of output to input force in simple machines.

Flashcard 24: State the formula for elastic potential energy stored in an ideal spring compressed or stretched by xx.

Answer: Us=12kx2U_s = \frac{1}{2}kx^2. Elastic potential energy derives from Hooke's law, integrating force over displacement to store energy quadratically with deformation.

Flashcard 25: State the ideal simple-machine energy relation between input and output work.

Answer: Win=WoutW_{in} = W_{out}. In ideal machines without friction, energy conservation implies equal input and output work despite force-distance trade-offs.