Physics Flashcards: Model Energy Transfer Computationally

Study Model Energy Transfer Computationally in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Physics

Model Energy Transfer Computationally

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What is the elastic potential energy formula for a spring in computational energy tracking?

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ANSWER

Us=12kx2U_s=\frac{1}{2}kx^2. Elastic PE is proportional to spring constant and displacement squared.

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Flashcard 1: What is the elastic potential energy formula for a spring in computational energy tracking?

Answer: Us=12kx2U_s=\frac{1}{2}kx^2. Elastic PE is proportional to spring constant and displacement squared.

Flashcard 2: What is the work done by a constant force parallel to the displacement?

Answer: W=FdW=Fd. Work equals force times distance when parallel.

Flashcard 3: What is the net work–kinetic energy relation used to compute speed changes?

Answer: Wnet=ΔKW_{\text{net}}=\Delta K. Work-energy theorem: net work changes kinetic energy.

Flashcard 4: What is the power definition used to model energy transfer rate?

Answer: P=ΔEΔtP=\frac{\Delta E}{\Delta t}. Power is energy transfer rate.

Flashcard 5: Find spring energy UsU_s for k=200N/mk=200\,\text{N/m} and x=0.10mx=0.10\,\text{m}.

Answer: 1J1\,\text{J}. Us=12(200)(0.1)2=12(200)(0.01)=1JU_s = \frac{1}{2}(200)(0.1)^2 = \frac{1}{2}(200)(0.01) = 1\,\text{J}

Flashcard 6: What is the work formula for a constant force parallel to displacement?

Answer: W=FdW=Fd. Work equals force times distance when force is parallel to motion.

Flashcard 7: What equation relates energy transferred by heating to mass, specific heat, and temperature change?

Answer: Q=mcΔTQ=mc\Delta T. Heat capacity relates thermal energy to temperature change.

Flashcard 8: What is the thermal energy transfer from kinetic friction over distance dd on a level surface?

Answer: Eth=fkd=μkNdE_{\text{th}}=f_k d=\mu_k N d. Friction converts mechanical energy to thermal over distance.

Flashcard 9: Find heat QQ for m=0.50kgm=0.50\,\text{kg}, c=4000\,\text{J/(kg\cdot ^\circ C)}, ΔT=2C\Delta T=2\,^\circ\text{C}.

Answer: 4000J4000\,\text{J}. Q=(0.50)(4000)(2)=4000JQ=(0.50)(4000)(2)=4000\,\text{J}.

Flashcard 10: Find work WW for F=10NF=10\,\text{N}, d=5md=5\,\text{m}, and θ=60\theta=60^\circ.

Answer: 25J25\,\text{J}. W=(10)(5)cos(60°)=50(0.5)=25JW = (10)(5)\cos(60°) = 50(0.5) = 25\,\text{J}

Flashcard 11: Find average power for ΔE=600J\Delta E=600\,\text{J} transferred in Δt=3s\Delta t=3\,\text{s}.

Answer: 200W200\,\text{W}. P=6003=200WP = \frac{600}{3} = 200\,\text{W}

Flashcard 12: What is the conservation of energy equation for a closed system in computational models?

Answer: Einitial=EfinalE_{\text{initial}}=E_{\text{final}}. Energy is conserved in closed systems with no external work or heat transfer.

Flashcard 13: What is the work done by a constant force at angle θ\theta to the displacement?

Answer: W=FdcosθW=Fd\cos\theta. Cosine accounts for the component of force along displacement.

Flashcard 14: What is the kinetic energy formula used in energy-transfer calculations?

Answer: K=12mv2K=\frac{1}{2}mv^2. Kinetic energy depends on mass and velocity squared.

Flashcard 15: Identify the computational update rule for speed from kinetic energy KK and mass mm.

Answer: v=2Kmv=\sqrt{\frac{2K}{m}}. Rearranged kinetic energy formula to solve for speed.

Flashcard 16: Find work WW for F=10NF=10\,\text{N}, d=4md=4\,\text{m}, and θ=60\theta=60^\circ.

Answer: 20J20\,\text{J}. W=(10)(4)cos(60)=(10)(4)(0.5)=20JW=(10)(4)\cos(60^\circ)=(10)(4)(0.5)=20\,\text{J}.

Flashcard 17: What is the thermal energy change formula for heating with no phase change?

Answer: Q=mcΔTQ=mc\Delta T. Heat transfer depends on mass, specific heat, and temperature change.

Flashcard 18: What is the work formula for a constant force at angle θ\theta to displacement?

Answer: W=FdcosθW=Fd\cos\theta. Work depends on the component of force in the direction of motion.

Flashcard 19: Identify the correct discrete update for speed from kinetic energy KK and mass mm.

Answer: v=2Kmv=\sqrt{\frac{2K}{m}}. Rearranging K=12mv2K=\frac{1}{2}mv^2 to solve for velocity.

Flashcard 20: What is the energy balance statement including nonconservative work WncW_{\text{nc}}?

Answer: ΔK+ΔU=Wnc\Delta K+\Delta U=W_{\text{nc}}. Total mechanical energy change equals work done by nonconservative forces.

Flashcard 21: What is the conservation of energy statement used in computational energy-transfer models?

Answer: EinEout=ΔEsystemE_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}. Energy conservation: input minus output equals system change.

Flashcard 22: Find efficiency η\eta if Ein=200JE_{\text{in}}=200\,\text{J} and Eout=150JE_{\text{out}}=150\,\text{J}.

Answer: 0.750.75. η=150200=0.75\eta=\frac{150}{200}=0.75 or 75% efficient.

Flashcard 23: What is the mechanical power relation connecting energy transfer and time?

Answer: P=ΔEΔtP=\frac{\Delta E}{\Delta t}. Power is the rate of energy transfer.

Flashcard 24: Find KK for m=2kgm=2\,\text{kg} and v=3m/sv=3\,\text{m/s} using K=12mv2K=\frac{1}{2}mv^2.

Answer: 9J9\,\text{J}. K=12(2)(32)=12(2)(9)=9JK = \frac{1}{2}(2)(3^2) = \frac{1}{2}(2)(9) = 9\,\text{J}

Flashcard 25: What is the efficiency formula for an energy-transfer process in computational models?

Answer: η=EoutEin\eta=\frac{E_{\text{out}}}{E_{\text{in}}}. Efficiency is the ratio of useful output to total input energy.

Flashcard 26: What is the gravitational potential energy function for two masses separated by distance rr?

Answer: U=Gm1m2rU=-\frac{Gm_1m_2}{r}. Negative sign indicates attractive force lowers PE as rr decreases.

Flashcard 27: Find ΔUg\Delta U_g for m=1kgm=1\,\text{kg}, Δh=5m\Delta h=5\,\text{m}, g=10m/s2g=10\,\text{m/s}^2.

Answer: 50J50\,\text{J}. ΔUg=(1)(10)(5)=50J\Delta U_g=(1)(10)(5)=50\,\text{J}.

Flashcard 28: Find ΔUg\Delta U_g for m=4kgm=4\,\text{kg}, Δh=2m\Delta h=2\,\text{m}, g=10m/s2g=10\,\text{m/s}^2.

Answer: 80J80\,\text{J}. ΔUg=(4)(10)(2)=80J\Delta U_g = (4)(10)(2) = 80\,\text{J}

Flashcard 29: Identify the correct discrete update for energy using constant power PP over time step Δt\Delta t.

Answer: En+1=En+PΔtE_{n+1}=E_n+P\Delta t. Energy at next time step equals current energy plus power times time interval.

Flashcard 30: What is the change in gravitational potential energy near Earth?

Answer: ΔUg=mgΔh\Delta U_g=mg\Delta h. Gravitational PE change is weight times height change.

Flashcard 31: What is the average power in terms of work done over a time interval?

Answer: P=WΔtP=\frac{W}{\Delta t}. Power equals work divided by time interval.

Flashcard 32: What equation updates system energy when heat QQ is added and work WW is done by the system?

Answer: ΔE=QW\Delta E=Q-W. First law of thermodynamics: energy change equals heat added minus work done by system.

Flashcard 33: What is the gravitational potential energy change near Earth used in energy models?

Answer: ΔUg=mgΔh\Delta U_g=mg\Delta h. Gravitational PE change is weight times height change.

Flashcard 34: Identify the computational update rule for energy using constant power over time step Δt\Delta t.

Answer: Enew=Eold+PΔtE_{\text{new}}=E_{\text{old}}+P\Delta t. Energy increases by power times elapsed time.

Flashcard 35: What is the elastic potential energy stored in an ideal spring compressed or stretched by xx?

Answer: Us=12kx2U_s=\frac{1}{2}kx^2. Spring PE is quadratic in displacement from equilibrium.