Physics Quiz: Analyze Energy Using Conservation Laws

What this quiz covers

This quiz focuses on Analyze Energy Using Conservation Laws, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 1.4 m/s
2. 2.0 m/s
3. 2.4 m/s (correct answer)
4. 3.5 m/s

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a spring-mass system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. When the spring is compressed by distance x=0.30 m, it stores elastic potential energy PE_spring = ½kx² = ½(200 N/m)(0.30 m)² = 9 J. When released on the frictionless surface, this potential energy converts to kinetic energy of the attached mass: just after leaving the spring (at equilibrium), KE = ½mv² = 9 J, which gives v = √(2 × 9 / 3) = √6 ≈ 2.45 m/s. Choice C is correct because it properly applies conservation of energy E_initial = E_final and correctly calculates using the formula v = x √(k/m) ≈ 2.4 m/s. Choice D forgets the ½ factor in the kinetic energy formula or misapplies the square root, calculating a higher speed that violates conservation by exceeding the initial spring energy. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Remember that mechanical energy (KE + PE) is only conserved when no non-conservative forces act—if friction, air resistance, or inelastic collisions occur, mechanical energy decreases as it converts to thermal energy, but total energy including thermal is always conserved: you can test this by calculating initial mechanical energy, final mechanical energy, and the work done by friction: E_mech-i = E_mech-f + W_friction should hold exactly, demonstrating that the 'lost' mechanical energy is accounted for as thermal energy.

Question 2

1. 5.0 m/s
2. 7.1 m/s (correct answer)
3. 10 m/s
4. 14 m/s

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a pendulum system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. At the highest point of the swing, the pendulum has maximum gravitational PE = mgh = (4.0 kg)(10 m/s²)(2.5 m) = 100 J and zero KE since released from rest, while at the lowest point it has minimum PE (zero) and maximum KE = ½mv². Energy conservation gives mgh_top = ½mv²_bottom, so v = √(2gh) = √(2 × 10 × 2.5) = √50 ≈ 7.07 m/s, and since ignoring air resistance, total mechanical energy remains constant. Choice B is correct because it properly applies conservation of energy E_initial = E_final and correctly calculates the speed using v = √(2gh) ≈ 7.1 m/s. Choice C confuses the calculation by omitting the ½ in KE or using v = √(gh), leading to √25 = 5 m/s which is too low, or possibly √(4gh) = 10 m/s which overestimates. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Remember that mechanical energy (KE + PE) is only conserved when no non-conservative forces act—if friction, air resistance, or inelastic collisions occur, mechanical energy decreases as it converts to thermal energy, but total energy including thermal is always conserved: you can test this by calculating initial mechanical energy, final mechanical energy, and the work done by friction: E_mech-i = E_mech-f + W_friction should hold exactly, demonstrating that the 'lost' mechanical energy is accounted for as thermal energy.

Question 3

1. $0\,\text{J}$
2. $4\,\text{J}$
3. $8\,\text{J}$ (correct answer)
4. $16\,\text{J}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. For pendulum/oscillation: At the highest point of the swing, the pendulum has maximum gravitational PE = mgh = (1 kg)(10 m/s²)(0.8 m) = 8 J (where h is height above lowest point) and minimum KE (zero if momentarily at rest), while at the lowest point it has minimum PE (zero if we use lowest point as reference) and maximum KE = ½mv². Energy conservation gives mgh_top = ½mv²_bottom, and if no friction acts, this total mechanical energy remains constant throughout the swing: E_mech = mgh + ½mv² = constant at every point, so KE_bottom = 8 J. Choice C is correct because it properly applies conservation of energy E_initial = E_final accounting for all forms and accurately identifies the energy transformation sequence. Choice D violates conservation of energy by claiming energy is destroyed rather than transformed—energy never disappears, it only changes form from mechanical (KE and PE) to thermal energy when friction acts, or between KE and PE when only conservative forces act. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 4

1. $96\,\text{J}$
2. $192\,\text{J}$ (correct answer)
3. $480\,\text{J}$
4. $0\,\text{J}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system with friction. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. Initially the sled has mechanical energy E_mech-i = PE_i + KE_i = mgh + 0 = (4 kg)(10 m/s²)(12 m) = 480 J, and finally it has E_mech-f = PE_f + KE_f = 0 + ½mv² = ½(4)(144) = 288 J. The decrease in mechanical energy ΔE_mech = 480 - 288 = 192 J equals the work done by friction, which converted this mechanical energy to thermal energy (heat) through the friction force acting over distance: the total energy is still conserved when we include thermal: E_mech-i = E_mech-f + E_thermal. Choice B is correct because it properly applies conservation of energy E_initial = E_final accounting for all forms and correctly calculates the thermal energy as the difference in mechanical energy. Choice C confuses the total initial potential energy with the thermal energy, claiming 480 J when actually only the portion not converted to kinetic energy becomes thermal: 480 J initial PE transforms to 288 J KE + 192 J thermal, not all to thermal. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 5

1. $2.0\,\text{m/s}$
2. $3.5\,\text{m/s}$ (correct answer)
3. $6.0\,\text{m/s}$
4. $12\,\text{m/s}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. When the spring is compressed by distance x = 0.30 m, it stores elastic potential energy PE_spring = ½kx² = ½(200 N/m)(0.30 m)² = 9 J. When released, this potential energy converts to kinetic energy of the attached mass: at maximum speed (passing through equilibrium), KE = ½mv² equals the initial PE_spring, so ½mv² = ½kx², which gives v = √(k/m) × x = √(200/1.5) × 0.3 ≈ 3.46 m/s, approximately 3.5 m/s. Choice B is correct because it properly applies conservation of energy E_initial = E_final accounting for all forms and correctly calculates using appropriate formula KE = ½mv² or PE_spring = ½kx². Choice C forgets to take the square root properly or doubles the value, perhaps calculating v = (k/m) x or similar error, leading to an overestimated speed that violates energy equivalence. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 6

1. $\sqrt{150}\,\text{m/s}$
2. $15\,\text{m/s}$
3. $30\,\text{m/s}$
4. $\sqrt{300}\,\text{m/s}$ (correct answer)

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. In this scenario, the object starts at height h = 15 m with gravitational potential energy PE = mgh = (2 kg)(10 m/s²)(15 m) = 300 J and zero kinetic energy (if starting from rest). As it falls, PE converts to KE, so at the bottom where h = 0, all the initial PE has become KE: ½mv² = mgh, solving for velocity gives v = √(2gh) = √(2 × 10 × 15) = √300 m/s, demonstrating complete PE → KE transformation. Choice D is correct because it properly applies conservation of energy E_initial = E_final accounting for all forms and correctly calculates using appropriate formula KE = ½mv² or PE = mgh. Choice C forgets the factor of 2 in the velocity formula, calculating v = √(gh) = √(150) instead of √(2gh) = √300, which underestimates the speed by a factor of √2. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 7

1. Mechanical energy was conserved; no energy changed form.
2. Some gravitational potential energy became kinetic energy and the rest became thermal energy due to friction. (correct answer)
3. Energy was destroyed by friction so total energy decreased.
4. Kinetic energy was converted into gravitational potential energy as she went downhill.

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system with friction. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. Initial PE = mgh = (5 kg)(10 m/s²)(8 m) = 400 J, final KE = ½mv² = ½(5)(100) = 250 J, so some PE converted to KE and the rest to thermal: E_thermal = 400 - 250 = 150 J due to rough hill. This shows partial PE → KE with friction causing PE → thermal. Choice B is correct because it accurately identifies the energy transformation sequence and correctly explains that decreased mechanical energy was converted to thermal energy, not destroyed. Choice C violates conservation of energy by claiming energy is destroyed rather than transformed—energy never disappears, it only changes form from mechanical (KE and PE) to thermal energy when friction acts, or between KE and PE when only conservative forces act. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 8

1. 3.6 m
2. 7.2 m (correct answer)
3. 12 m
4. 14.4 m

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so $E_initial = E_final$ when all energy forms are accounted for. Starting at the bottom with speed $v = 12 \text{ m/s}$, the skateboarder has kinetic energy $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1.0 \text{ kg})(12 \text{ m/s})^2 = \frac{1}{2}(1.0)(144) = 72 \text{ J}$ and zero potential energy (bottom is reference). At maximum height where the skateboarder momentarily stops, all kinetic energy has converted to gravitational potential energy: $KE_bottom = PE_top$, so $72 \text{ J} = mgh = (1.0 \text{ kg})(10 \text{ m/s}^2)h$, solving for $h$: $h = 72/10 = 7.2 \text{ m}$. Choice B is correct because it properly applies conservation of energy, recognizing that all $72 \text{ J}$ of initial kinetic energy transforms into gravitational potential energy at maximum height where $v = 0$. Choice A ($3.6 \text{ m}$) would only account for $36 \text{ J}$ of energy, exactly half the initial kinetic energy, possibly from using $KE = \frac{1}{2}mv^2$ but forgetting the $\frac{1}{2}$ when setting up the energy equation. Choice C ($12 \text{ m}$) might result from confusing the initial speed ($12 \text{ m/s}$) with the height, while choice D ($14.4 \text{ m}$) would require $144 \text{ J}$ of initial energy, possibly from calculating $KE = mv^2$ instead of $KE = \frac{1}{2}mv^2$. When solving conservation of energy problems: (1) identify all energy forms present initially and finally ($KE$, $PE_g$, $PE_spring$, thermal), (2) write $E_initial = E_final$ explicitly: for conservative forces only, use $KE_i + PE_i = KE_f + PE_f$; for friction present, use $KE_i + PE_i = KE_f + PE_f + E_thermal$, (3) substitute formulas: $KE = \frac{1}{2}mv^2$, $PE_g = mgh$, $PE_spring = \frac{1}{2}kx^2$, (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.).

Question 9

1. 0.25 m
2. 0.50 m
3. 1.25 m (correct answer)
4. 2.50 m

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. When the spring is compressed by distance x = 0.20 m, it stores elastic potential energy PE_spring = ½kx² = ½(250 N/m)(0.20 m)² = ½(250)(0.04) = 5 J. When released, this spring potential energy converts entirely to gravitational potential energy at the maximum height: PE_spring = PE_gravity, so ½kx² = mgh, which gives 5 J = (0.40 kg)(10 m/s²)h, solving for h: h = 5/(0.40 × 10) = 5/4 = 1.25 m. Choice C is correct because it properly applies conservation of energy, recognizing that all 5 J of spring potential energy transforms into gravitational potential energy at maximum height. Choice B (0.50 m) would only account for 2 J of energy (mgh = 0.40 × 10 × 0.50 = 2 J), while choice D (2.50 m) would require 10 J of initial energy, twice what the spring actually stores. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Key energy transformation patterns to recognize: falling objects convert PE → KE, rising objects convert KE → PE, springs oscillate between elastic PE ↔ KE, friction always converts mechanical energy (KE + PE) → thermal energy, and in all cases total energy remains constant when all forms are included.

Question 10

1. 6 J
2. 30 J
3. 60 J (correct answer)
4. 120 J

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In this scenario, the ball starts at height h = 12 m with gravitational potential energy PE = mgh = (0.50 kg)(10 m/s²)(12 m) = 60 J and zero kinetic energy (dropped from rest). As it falls, PE converts to KE, so at the bottom where h = 0, all the initial PE has become KE: KE_final = PE_initial = 60 J, demonstrating complete PE → KE transformation. Choice C is correct because it properly applies conservation of energy E_initial = E_final, recognizing that all 60 J of initial potential energy transforms into kinetic energy at the bottom. Choice B (30 J) incorrectly suggests only half the potential energy converts to kinetic, which would violate conservation of energy in a frictionless system, while choice D (120 J) doubles the correct value, perhaps by using KE = mv² instead of KE = ½mv². When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Key energy transformation patterns to recognize: falling objects convert PE → KE, rising objects convert KE → PE, springs oscillate between elastic PE ↔ KE, friction always converts mechanical energy (KE + PE) → thermal energy, and in all cases total energy remains constant when all forms are included.

Question 11

1. 5.0 m/s
2. 10 m/s (correct answer)
3. 20 m/s
4. 50 m/s

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. At the highest point of the swing, the pendulum has maximum gravitational PE = mgh = (1.0 kg)(10 m/s²)(5.0 m) = 50 J and minimum KE (zero since released from rest), while at the lowest point it has minimum PE (zero if we use lowest point as reference) and maximum KE = ½mv². Energy conservation gives mgh_top = ½mv²_bottom, so 50 J = ½(1.0 kg)v², solving for v: v² = 100, therefore v = 10 m/s, demonstrating complete PE → KE transformation. Choice B is correct because it properly applies conservation of energy E_initial = E_final, correctly calculating v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. Choice A (5.0 m/s) forgets the factor of 2 in the formula v = √(2gh), calculating v = √(gh) = √50 ≈ 7.1 m/s then rounding down, while choice C (20 m/s) would require the pendulum to start from a height of 20 m, not 5 m. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Remember that for a pendulum with no friction, mechanical energy oscillates between PE at the extremes and KE at the bottom, but the total mechanical energy E = PE + KE remains constant throughout the swing.

Question 12

1. $32\,\text{J}$ (correct answer)
2. $48\,\text{J}$
3. $80\,\text{J}$
4. $128\,\text{J}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. Initially the ball has mechanical energy E_mech-i = PE_i = mgh_i = (0.80 kg)(10 m/s²)(10 m) = 80 J (dropped from rest, so KE_i = 0). After the bounce, it reaches maximum height h_f = 6.0 m with E_mech-f = PE_f = mgh_f = (0.80 kg)(10 m/s²)(6.0 m) = 48 J. The decrease in mechanical energy ΔE_mech = E_mech-i - E_mech-f = 80 J - 48 J = 32 J equals the energy converted to thermal and sound during the inelastic bounce. Choice A (32 J) is correct because it accurately calculates the difference between initial mechanical energy (80 J) and final mechanical energy (48 J), recognizing this 32 J was converted to thermal/sound energy during the bounce. Choice C (80 J) incorrectly claims all the initial energy was lost, ignoring that the ball still bounced to 6.0 m height, retaining 48 J of mechanical energy. When analyzing bouncing problems: (1) calculate initial mechanical energy before the bounce, (2) calculate final mechanical energy after the bounce (at maximum rebound height), (3) the difference represents energy converted to thermal/sound: E_lost = mg(h_initial - h_final), and (4) verify that this "lost" energy plus final mechanical energy equals initial energy. Remember that real bounces are never perfectly elastic—some mechanical energy always converts to thermal energy and sound, which is why bouncing objects don't return to their original height.

Question 13

1. $7.2\,\text{m}$
2. $9.2\,\text{m}$ (correct answer)
3. $14\,\text{m}$
4. $16\,\text{m}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. Initially, the ball has both kinetic and potential energy: KE_i = ½mv² = ½(1.2 kg)(12 m/s)² = 86.4 J and PE_i = mgh_i = (1.2 kg)(10 m/s²)(2.0 m) = 24 J, giving total energy E_total = 110.4 J. At maximum height, all energy is potential (KE = 0), so PE_max = mgh_max = 110.4 J, which gives h_max = 110.4/(1.2 × 10) = 9.2 m above ground. Choice B (9.2 m) is correct because it properly applies conservation of energy, accounting for both initial kinetic energy (86.4 J) and initial potential energy (24 J), recognizing that at maximum height all 110.4 J becomes gravitational PE. Choice A (7.2 m) incorrectly calculates only the height gained from the initial kinetic energy (86.4 J ÷ 12 = 7.2 m), forgetting to add the initial 2.0 m platform height. When solving projectile problems with initial height: (1) calculate total initial energy E_i = KE_i + PE_i = ½mv² + mgh_platform, (2) at maximum height, all energy is potential: E_total = mgh_max, (3) solve for maximum height above ground, and (4) verify by checking that the height gained above the platform equals v²/(2g). Alternative approach: the ball rises Δh = v²/(2g) = 144/20 = 7.2 m above the platform, so total height above ground is 2.0 + 7.2 = 9.2 m, confirming our energy calculation.

Question 14

1. $10\,\text{m/s}$
2. $14\,\text{m/s}$ (correct answer)
3. $20\,\text{m/s}$
4. $5\,\text{m/s}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In this scenario, the cart starts at height h = 10 m with gravitational potential energy PE = mgh = (2.0 kg)(10 m/s²)(10 m) = 200 J and zero kinetic energy (starting from rest). As it falls on the frictionless track, PE converts to KE, so at the bottom where h = 0, all the initial PE has become KE: ½mv² = mgh, solving for velocity gives v = √(2gh) = √(2 × 10 × 10) = √200 = 14.1 m/s, demonstrating complete PE → KE transformation. Choice B (14 m/s) is correct because it properly applies conservation of energy E_initial = E_final, correctly calculating that all 200 J of initial PE converts to KE at the bottom, giving v = 14 m/s. Choice C (20 m/s) forgets the ½ factor in the kinetic energy formula, calculating KE = mv² instead of KE = ½mv², which would require 400 J of energy instead of the available 200 J, violating conservation. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f, (3) substitute formulas: KE = ½mv², PE_g = mgh, (4) solve for the unknown, and (5) verify your answer makes physical sense. Key energy transformation patterns to recognize: falling objects convert PE → KE, rising objects convert KE → PE, and in frictionless systems mechanical energy (KE + PE) is conserved throughout the motion.

Question 15

1. $0.60\,\text{m}$
2. $1.2\,\text{m}$
3. $2.4\,\text{m}$ (correct answer)
4. $4.8\,\text{m}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. When the spring is compressed by distance x = 0.40 m, it stores elastic potential energy PE_spring = ½kx² = ½(300 N/m)(0.40 m)² = ½(300)(0.16) = 24 J. After leaving the spring, this energy converts entirely to gravitational potential energy at maximum height: PE_g = mgh = 24 J, so (1.0 kg)(10 m/s²)h = 24 J, which gives h = 2.4 m. Choice C (2.4 m) is correct because it properly applies conservation of energy, recognizing that all 24 J of initial spring potential energy converts to gravitational potential energy at maximum height, giving h = 2.4 m. Choice B (1.2 m) incorrectly calculates only half the correct height, possibly by forgetting the ½ factor in the spring potential energy formula and using PE = kx² = 48 J, then getting h = 4.8 m, or by making another calculation error. When solving multi-stage energy problems: (1) calculate initial energy in the spring PE_spring = ½kx², (2) recognize that at maximum height all energy is gravitational PE (KE = 0 at the peak), (3) set PE_spring = PE_g and solve: ½kx² = mgh, so h = kx²/(2mg), and (4) verify the answer is reasonable. Key insight: in vertical launches without friction, the initial elastic potential energy in the spring equals the gravitational potential energy at maximum height—the block momentarily stops at the peak, converting all energy to PE_g.

Question 16

1. $20\,\text{m/s}$
2. $14\,\text{m/s}$
3. $11\,\text{m/s}$
4. $16\,\text{m/s}$ (correct answer)

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. Initially the cart has mechanical energy E_mech-i = PE_i + KE_i = mgh + 0 = (1.5 kg)(10 m/s²)(20 m) = 300 J, and finally it has E_mech-f = PE_f + KE_f = 0 + ½mv². The decrease in mechanical energy equals the thermal energy created by friction: 90 J of mechanical energy was converted to thermal energy, so the final kinetic energy is KE_f = 300 J - 90 J = 210 J. Choice D is correct because it properly applies conservation of energy accounting for thermal energy: ½(1.5)v² = 210, which gives v² = 280, so v = 16.7 m/s ≈ 16 m/s, correctly showing that the cart retains 210 J as kinetic energy after losing 90 J to friction. Choice A (20 m/s) incorrectly assumes no energy loss to friction, calculating v = √(2gh) = √(400) = 20 m/s, which would only be correct if the track were frictionless and all PE converted to KE. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Remember that mechanical energy (KE + PE) is only conserved when no non-conservative forces act—if friction, air resistance, or inelastic collisions occur, mechanical energy decreases as it converts to thermal energy, but total energy including thermal is always conserved: you can test this by calculating initial mechanical energy (300 J), final mechanical energy (210 J), and the thermal energy (90 J): 300 J = 210 J + 90 J holds exactly.

Question 17

1. $4.2\,\text{m/s}$
2. $6.0\,\text{m/s}$ (correct answer)
3. $8.5\,\text{m/s}$
4. $12\,\text{m/s}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. When the spring is compressed by distance x = 0.30 m, it stores elastic potential energy PE_spring = ½kx² = ½(200 N/m)(0.30 m)² = ½(200)(0.09) = 9.0 J. When released, this potential energy converts to kinetic energy of the attached mass: at maximum speed (when the spring returns to natural length), KE = ½mv² equals the initial PE_spring, so ½(0.50)v² = 9.0, which gives v² = 36, so v = 6.0 m/s. Choice B is correct because it properly applies conservation of energy: the spring's initial elastic potential energy of 9.0 J completely converts to the block's kinetic energy as it leaves the spring, giving the block a speed of 6.0 m/s. Choice D (12 m/s) incorrectly doubles the speed, perhaps by forgetting the ½ factor in either the spring potential energy or kinetic energy formula, which would make the energy calculation incorrect and violate conservation. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Remember that mechanical energy (KE + PE) is only conserved when no non-conservative forces act—in this frictionless case, the elastic PE of the spring transforms completely into KE of the block.

Question 18

1. $18\,\text{J}$
2. $90\,\text{J}$
3. $180\,\text{J}$ (correct answer)
4. $360\,\text{J}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In this scenario, the ball starts at height h = 12 m with gravitational potential energy PE = mgh = (1.5 kg)(10 m/s²)(12 m) = 180 J and zero kinetic energy (dropped from rest). As it falls, PE converts to KE, so just before hitting the ground where h = 0, all the initial PE has become KE: KE_final = PE_initial = 180 J, demonstrating complete PE → KE transformation. Choice C is correct because it properly applies conservation of energy, recognizing that all 180 J of initial potential energy converts to kinetic energy at the bottom. Choice A (18 J) appears to forget the factor of 10 in g, calculating as if g = 1 m/s², while choice B (90 J) is exactly half the correct answer, possibly from using PE = ½mgh incorrectly, and choice D (360 J) doubles the correct answer, violating conservation by creating energy from nowhere. When solving conservation of energy problems: (1) identify all energy forms present initially and finally, (2) write E_initial = E_final explicitly: for free fall with no air resistance, use PE_i + KE_i = PE_f + KE_f, (3) substitute values: here PE_i = mgh = 180 J and KE_i = 0, while PE_f = 0 and KE_f = ?, (4) solve: KE_f = 180 J. Key energy transformation patterns to recognize: falling objects convert PE → KE completely when no friction acts, and the kinetic energy gained exactly equals the potential energy lost—this direct conversion makes free fall problems particularly straightforward.

Question 19

1. $mgh = \tfrac{1}{2}mv^2 - E_{\text{thermal}}$
2. $mgh = \tfrac{1}{2}mv^2 + E_{\text{thermal}}$ (correct answer)
3. $\tfrac{1}{2}mv^2 = mgh + E_{\text{thermal}}$
4. $mgh + E_{\text{thermal}} = \tfrac{1}{2}mv^2$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations in a system. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. Initially at height h = 10 m, the car has PE = mgh = (1.0 kg)(10 m/s²)(10 m) = 100 J and KE = 0 (starting from rest). At the bottom, it has KE = ½mv² = ½(1.0)(10)² = 50 J and PE = 0. Since only 50 J appears as kinetic energy but 100 J of potential energy was available, the "missing" 50 J was converted to thermal energy by friction: E_thermal = 100 J - 50 J = 50 J. Choice B correctly states mgh = ½mv² + E_thermal, which rearranges to 100 = 50 + 50, properly accounting for the initial PE equaling the sum of final KE plus thermal energy. Choice A incorrectly subtracts thermal energy, suggesting energy is destroyed rather than conserved; choice C incorrectly puts KE on the left side as if it were the initial energy; choice D suggests thermal energy was present initially, which contradicts the physics of friction converting mechanical to thermal energy during motion. When writing conservation equations with friction: start with E_initial = E_final, identify all energy forms at each stage, and remember that friction converts mechanical energy TO thermal (never FROM thermal). The general form is (KE + PE)_initial = (KE + PE)_final + E_thermal, which for this problem simplifies to mgh = ½mv² + E_thermal since initial KE and final PE are both zero.

Question 20

1. $mgh=\tfrac12 mv^2$
2. $mgh+E_{thermal}=\tfrac12 mv^2$
3. $mgh=\tfrac12 mv^2+E_{thermal}$ (correct answer)
4. $\tfrac12 mv^2=mgh+E_{thermal}$

Explanation: This question tests understanding of conservation of energy and the ability to analyze energy transformations including thermal energy from friction. The law of conservation of energy states that the total energy in an isolated system remains constant—energy can transform between different forms (kinetic, potential, thermal, etc.) but cannot be created or destroyed, so E_initial = E_final when all energy forms are accounted for. In systems with only conservative forces like gravity and springs, mechanical energy (KE + PE) is conserved, but when non-conservative forces like friction act, mechanical energy decreases as it converts to thermal energy, though total energy including thermal is still conserved. Initial PE = mgh = (5 kg)(10 m/s²)(20 m) = 1000 J, final KE = ½mv² = ½(5)(16)² = 640 J, so E_thermal = 1000 - 640 = 360 J, showing some PE converted to KE and some to thermal. The equation mgh = ½mv² + E_thermal correctly accounts for all forms. Choice C is correct because it properly applies conservation of energy E_initial = E_final accounting for all forms, with initial PE equaling final KE plus thermal energy. Choice B incorrectly reverses the transformation direction, stating mgh + E_thermal = ½mv², which would imply thermal energy adds to initial rather than accounting for the loss. When solving conservation of energy problems: (1) identify all energy forms present initially and finally (KE, PE_g, PE_spring, thermal), (2) write E_initial = E_final explicitly: for conservative forces only, use KE_i + PE_i = KE_f + PE_f; for friction present, use KE_i + PE_i = KE_f + PE_f + E_thermal, (3) substitute formulas: KE = ½mv², PE_g = mgh, PE_spring = ½kx², (4) solve for the unknown, and (5) verify your answer makes physical sense (velocities reasonable, energies positive, etc.). Key energy transformation patterns to recognize: falling objects convert PE → KE, rising objects convert KE → PE, springs oscillate between elastic PE ↔ KE, friction always converts mechanical energy (KE + PE) → thermal energy, and in all cases total energy remains constant when all forms are included—the phrase 'missing energy' means energy that transformed from one form to another, not energy that disappeared.