Model Energy Transfer Computationally - Grade 11 Physics

1. How far is the spring compressed?
   Answer: $1.73\ \text{m}$ When a roller coaster interacts with a spring, you're dealing with energy conservation across the entire system. The key is accounting for every form of energy: kinetic, gravitational potential, elastic potential, and thermal losses from friction. Start by finding the total mechanical energy at the high point: $KE = \frac{1}{2}(500)(5.0)^2 = 6{,}250\ \text{J}$ and $PE = (500)(9.8)(20.0) = 98{,}000\ \text{J}$, giving a total of $104{,}250\ \text{J}$. At the spring's location (5.0 m high), the gravitational PE is $(500)(9.8)(5.0) = 24{,}500\ \text{J}$. After subtracting friction losses of $50{,}000\ \text{J}$, the energy available to compress the spring is: $104{,}250 - 24{,}500 - 50{,}000 = 29{,}750\ \text{J}$. Setting this equal to elastic potential energy: $\frac{1}{2}(20{,}000)x^2 = 29{,}750$, so $x^2 = \frac{29{,}750}{10{,}000} = 2.975$, giving $x \approx 1.73\ \text{m}$. That confirms C. Choice A (1.22 m) results from forgetting to include the initial kinetic energy — a common mistake when students focus only on height. Choice B (1.53 m) comes from neglecting the gravitational PE at the final position (5.0 m), treating ground level as the reference incorrectly. Choice D (2.82 m) comes from omitting the friction term entirely, leaving too much energy for the spring. A great strategy here: always write out a full energy accounting equation before solving — $E_{\text{initial}} = E_{\text{final}} + E_{\text{lost}}$. Missed terms are the #1 source of errors on energy conservation problems.
2. Assuming the metal and water reach the same final temperature, what is that temperature?
   Answer: $27.9^\circ\text{C}$ When you see a heat transfer problem with an efficiency factor, the key is applying that percentage before setting up your energy balance. The framework here is calorimetry: heat lost by the hot object equals heat gained by the cool object — but only the portion that actually reaches the water counts. The metal releases total energy $Q_{metal} = mc\Delta T = (0.400)(450)(150 - T_f)$. Since only 90% reaches the water, the energy absorbed by water is $0.90 \times Q_{metal}$. Setting that equal to the water's heat gain: $0.90 \times (0.400)(450)(150 - T_f) = (0.600)(4200)(T_f - 20.0)$ $162(150 - T_f) = 2520(T_f - 20.0)$ $24300 - 162T_f = 2520T_f - 50400$ $74700 = 2682T_f$ $T_f \approx 27.9^\circ\text{C}$ That confirms A is correct. Choice B ($28.4^\circ\text{C}$) results from using approximately 92–93% efficiency instead of 90% — a small arithmetic slip in applying the percentage. Choice C ($28.7^\circ\text{C}$) comes from rounding intermediate values too aggressively before completing the algebra. Choice D ($70.9^\circ\text{C}$) is the classic trap: ignoring the 90% efficiency entirely and assuming 100% of the metal's heat reaches the water, which gives an unrealistically high equilibrium temperature. Your study tip: whenever a problem states that only a fraction of energy is transferred, apply that multiplier to the losing side of the equation before balancing — not to the final temperature directly.
3. Approximately how much sound energy passes through the detector in $20\ \text{s}$?
   Answer: $0.50\ \text{J}$ When a question involves sound spreading from a source, think in terms of intensity — how much power passes through each square meter at a given distance. The key chain is: electrical power → sound power → intensity at distance → power through detector → energy over time. Start by finding the sound power the speaker actually produces: $P_{sound} = 0.25 \times 40\ \text{W} = 10\ \text{W}$. This sound spreads uniformly over a sphere of radius $r = 4.0\ \text{m}$, so the surface area is $A_{sphere} = 4\pi (4.0)^2 \approx 201\ \text{m}^2$. The intensity (power per unit area) at the detector is $I = \frac{10}{201} \approx 0.0497\ \text{W/m}^2$. The power intercepted by the detector is $P_{det} = I \times A_{det} = 0.0497 \times 0.50 \approx 0.0249\ \text{W}$. Over $t = 20\ \text{s}$, the energy is $E = P_{det} \times t \approx 0.0249 \times 20 \approx 0.50\ \text{J}$, confirming B. Choice A ($0.025\ \text{J}$) results from forgetting to multiply by the 20 seconds — stopping at power rather than converting to energy. Choice C ($2.0\ \text{J}$) comes from using the full 40 W instead of applying the 25% efficiency, overstating the sound output by a factor of 4. Choice D ($8.0\ \text{J}$) skips the sphere-area step entirely, as if all 10 W passed through the detector — ignoring that the sound spreads over hundreds of square meters. Your strategy: chain the steps — efficiency → intensity → intercepted power → energy. Write each step explicitly; most errors happen when students skip or conflate one link in the chain.
4. What average electrical input power is required during the interval?
   Answer: $3.05\ \text{kW}$ When a question involves a motor lifting an object, you need to account for two forms of energy: the gravitational potential energy gained and the kinetic energy gained. Then, motor efficiency tells you how much extra electrical energy is needed to deliver that mechanical output. Start by calculating the total mechanical energy the motor must provide. The gravitational potential energy gained is $W_{grav} = mgh = 600 \times 9.8 \times 12.0 = 70{,}560\ \text{J}$. The elevator also accelerates from rest to $3.0\ \text{m/s}$, so kinetic energy gained is $W_{KE} = \frac{1}{2}mv^2 = \frac{1}{2}(600)(3.0)^2 = 2{,}700\ \text{J}$. Total mechanical output energy: $70{,}560 + 2{,}700 = 73{,}260\ \text{J}$. Since the motor is only 80% efficient, the electrical energy input must be larger: $E_{input} = \frac{73{,}260}{0.80} = 91{,}575\ \text{J}$. Dividing by the time interval gives average electrical power: $P = \frac{91{,}575}{30.0} \approx 3{,}053\ \text{W} \approx 3.05\ \text{kW}$, confirming D. Choice A (1.95 kW) results from forgetting both the kinetic energy term and the efficiency factor. Choice B (2.35 kW) comes from applying efficiency but forgetting kinetic energy entirely — only using gravitational PE. Choice C (2.44 kW) includes kinetic energy but forgets to divide by efficiency, giving the output power rather than the input power. A reliable strategy: always list every energy type (PE, KE, heat, etc.), sum them, then adjust for efficiency. Efficiency problems trip students up because it's easy to stop one step too early.
5. How long does the charging take, and how much energy is transferred to thermal energy in the charger?
   Answer: $2.00\ \text{h}$ and $80\ \text{W}\cdot\text{h}$ When a question combines battery capacity, charger efficiency, and energy dissipation, break it into two independent calculations: time and wasted energy. Finding the time: The battery needs to charge from 20% to 90%, so it gains 70% of its 50 A·h capacity: $0.70 \times 50 = 35\ \text{A}\cdot\text{h}$ of charge. The energy stored in the battery is $Q \times V = 35\ \text{A}\cdot\text{h} \times 12\ \text{V} = 420\ \text{W}\cdot\text{h}$. Since the charger is 84% efficient, the wall must supply $\frac{420}{0.84} = 500\ \text{W}\cdot\text{h}$ total. With the charger drawing 250 W from the wall, the time required is $\frac{500\ \text{W}\cdot\text{h}}{250\ \text{W}} = 2.00\ \text{h}$. That confirms D. Finding thermal energy: The charger wastes $100\% - 84\% = 16\%$ of the input energy as heat. Total wall energy used is $250\ \text{W} \times 2.00\ \text{h} = 500\ \text{W}\cdot\text{h}$, so thermal energy $= 0.16 \times 500 = 80\ \text{W}\cdot\text{h}$. Again, D. Choice A gets 80 W·h correct but uses an incorrect time of 1.68 h — this comes from forgetting to divide by efficiency when calculating wall energy needed. Choice B arrives at the right time but reports 420 W·h as thermal energy, confusing the energy stored in the battery with the energy lost as heat. Choice C uses 2.38 h, likely from dividing 500 W·h by 210 W instead of 250 W. A useful habit: always track energy flow — wall input splits into battery storage plus heat. Draw that flow explicitly so you never mix up which portion you're calculating.
6. How much kinetic energy is transferred to sound, thermal energy, and deformation, and what percentage of the initial kinetic energy does this represent?
   Answer: $96\ \text{kJ}$ and $40\%$ When you see a perfectly inelastic collision (objects lock together), your job is usually twofold: apply conservation of momentum to find the post-collision velocity, then compare kinetic energies before and after. The "lost" kinetic energy doesn't vanish — it converts into sound, heat, and deformation. Step 1 — Find post-collision velocity using momentum conservation: $p_i = p_f \implies (1200)(20.0) = (1200 + 800)v_f$ $v_f = \frac{24{,}000}{2000} = 12.0\ \text{m/s}$ Step 2 — Calculate initial and final kinetic energy: $KE_i = \frac{1}{2}(1200)(20.0)^2 = 240{,}000\ \text{J} = 240\ \text{kJ}$ $KE_f = \frac{1}{2}(2000)(12.0)^2 = 144{,}000\ \text{J} = 144\ \text{kJ}$ Step 3 — Find energy lost and percentage: $\Delta KE = 240 - 144 = 96\ \text{kJ}$ $\% = \frac{96}{240} \times 100 = 40\%$ This confirms answer B. Choice A ($86\ \text{kJ}$, $36\%$) comes from arithmetic errors, likely using incorrect masses or velocities in one of the steps. Choice C ($144\ \text{kJ}$, $60\%$) is a classic trap — students accidentally report the remaining kinetic energy rather than the lost kinetic energy. Choice D ($240\ \text{kJ}$, $100\%$) would only apply if both objects came to a complete stop, which violates momentum conservation here. A reliable strategy: always label $KE_i$, $KE_f$, and $\Delta KE$ separately. The question might ask for any one of the three, and mixing them up is the most common source of errors on inelastic collision problems.
7. What is the magnitude of the average air-resistance force during the upward motion?
   Answer: $0.74\ \text{N}$ Whenever a problem involves motion with friction or air resistance, the work-energy theorem is your most powerful tool: the net work done on an object equals its change in kinetic energy, $W_{net} = \Delta KE$. Here, the ball starts with $KE_i = \frac{1}{2}(0.200)(18.0)^2 = 32.4\ \text{J}$ and ends at rest at maximum height, so $\Delta KE = 0 - 32.4 = -32.4\ \text{J}$. Two forces do work during the upward trip: gravity and air resistance (both opposing the motion). Gravity does $W_g = -mgh = -(0.200)(9.8)(12.0) = -23.52\ \text{J}$. Setting up the equation: $W_g + W_{air} = \Delta KE$, so $-23.52 + W_{air} = -32.4$, giving $W_{air} = -8.88\ \text{J}$. Since $W_{air} = -F_{air} \cdot d$, the magnitude is $F_{air} = \frac{8.88}{12.0} \approx 0.74\ \text{N}$, confirming A is correct. Choice B ($0.98\ \text{N}$) equals exactly $mg/2$, suggesting a student guessed based on a fraction of gravity rather than using energy methods. Choice C ($1.23\ \text{N}$) likely results from sign errors — perhaps subtracting gravity's work in the wrong direction, inflating the air-resistance work. Choice D ($1.96\ \text{N}$) equals $mg$, a trap for students who confuse air resistance with the weight of the object entirely. Your study tip: when a problem mentions both gravity and a resistive force over a known distance, always reach for the work-energy theorem — it handles multiple forces cleanly and avoids the complications of Newton's second law with non-constant forces.
8. Ignoring evaporation, what is the final temperature of the water?
   Answer: $26.6^\circ\text{C}$ When a question combines electrical energy with heat transfer, your job is to build a chain: electrical energy → useful thermal energy → temperature change. Don't skip steps. Start by finding the total electrical energy: $E = VIt = 24 \times 5.0 \times (8.0 \times 60) = 57{,}600\ \text{J}$. Since only 75% reaches the water, the useful energy is $Q = 0.75 \times 57{,}600 = 43{,}200\ \text{J}$. Now apply the specific heat equation $Q = mc\Delta T$, solving for the temperature rise: $\Delta T = \frac{Q}{mc} = \frac{43{,}200}{1.20 \times 4200} = \frac{43{,}200}{5040} \approx 8.57^\circ\text{C}$. The final temperature is therefore $18.0 + 8.57 \approx 26.6^\circ\text{C}$, confirming D. Choice A ($18.1^\circ\text{C}$) results from a severe unit error — likely using seconds as minutes or forgetting to multiply by time entirely, producing a near-zero temperature rise. Choice B ($24.4^\circ\text{C}$) likely comes from forgetting the 75% efficiency factor, using only a fraction of the correct energy, and applying $\Delta T$ to the wrong base. Choice C ($29.4^\circ\text{C}$) is the trap for students who correctly calculate $\Delta T \approx 11.43^\circ\text{C}$ by skipping the 75% efficiency step entirely and using the full 57,600 J, then adding it to 18.0°C. A good habit: always list your chain of conversions before calculating — electrical energy, then efficiency loss, then heat equation. Efficiency factors are the most commonly missed step on energy-transfer questions.
9. What speed can the car reach using only the recovered energy?
   Answer: $16.5\ \text{m/s}$ When you see a question involving energy transfers with percentage efficiencies, treat it as a chain: calculate the starting energy, apply each efficiency multiplier in sequence, then use the final energy to find the unknown quantity. Start with the decrease in kinetic energy during braking: $\Delta KE = \frac{1}{2}(1400)(25.0^2 - 10.0^2) = \frac{1}{2}(1400)(625 - 100) = 367{,}500\ \text{J}$ The regenerative system stores 65% of this: $E_{\text{stored}} = 0.65 \times 367{,}500 = 238{,}875\ \text{J}$ The motor then converts 80% of that stored energy into kinetic energy: $KE_{\text{final}} = 0.80 \times 238{,}875 = 191{,}100\ \text{J}$ Now solve for the final speed using $KE = \frac{1}{2}mv^2$: $v = \sqrt{\frac{2 \times 191{,}100}{1400}} = \sqrt{273} \approx 16.5\ \text{m/s}$ This confirms C is correct. Choice A (10.8 m/s) results from applying only one of the two efficiency factors — typically just 65% — before solving for speed, missing the second conversion step. Choice B (18.5 m/s) comes from applying only the 80% motor efficiency to the total kinetic energy loss, skipping the 65% storage step entirely. Choice D (20.5 m/s) likely comes from forgetting to apply any efficiency factor and simply using the full $\Delta KE$ to find speed. A useful strategy: whenever a problem involves multiple energy conversions, write out each step explicitly and apply efficiencies one at a time. Skipping even one multiplier will land you on a trap answer.
10. What is the sled's speed at the bottom of the hill?
    Answer: $7.7\ \text{m/s}$ Whenever you see a sled (or any object) moving with friction on a slope, reach for the energy conservation equation with losses: the total mechanical energy at the top equals the total mechanical energy at the bottom plus energy lost to friction. $KE_{top} + PE_{top} = KE_{bottom} + E_{friction}$ Start by calculating each term. The initial kinetic energy is $KE_{top} = \frac{1}{2}(20.0)(4.0)^2 = 160\ \text{J}$. The gravitational potential energy lost during descent is $PE = mgh = (20.0)(9.8)(3.0) = 588\ \text{J}$. Friction removes $150\ \text{J}$, so the kinetic energy at the bottom is: $KE_{bottom} = 160 + 588 - 150 = 598\ \text{J}$ Solving for speed: $v = \sqrt{\frac{2(598)}{20.0}} = \sqrt{59.8} \approx 7.7\ \text{m/s}$, confirming B. Choice A (6.6 m/s) results from forgetting to include the initial kinetic energy — treating the sled as starting from rest. Choice C (8.6 m/s) comes from ignoring friction entirely, using only $KE_{top} + PE$ to find the final speed. Choice D (9.5 m/s) likely arises from both ignoring friction and adding extra energy incorrectly, or misapplying the formula altogether. A reliable strategy: always write out $E_{initial} = E_{final} + E_{lost}$ before plugging in numbers. This forces you to account for every energy term — initial KE, gravitational PE, and friction losses — so nothing gets accidentally dropped.

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