Potential Energy - College Physics

1. A 0.3 kg ball is dropped from rest at a height of 8.0 m above the ground. When the ball has fallen to a height of 3.2 m above the ground, what is its gravitational potential energy relative to the ground?
   Answer: 9.41 J When you encounter gravitational potential energy problems, remember that potential energy depends only on an object's current position, not its motion or starting point. The formula is $PE = mgh$, where m is mass, g is gravitational acceleration, and h is height above your reference point. Here, you need the potential energy when the ball is 3.2 m above the ground (your reference point). Simply substitute the values: $PE = (0.3 \text{ kg})(9.8 \text{ m/s}^2)(3.2 \text{ m}) = 9.41 \text{ J}$. This confirms answer A is correct. The wrong answers likely come from common calculation errors or conceptual mistakes. Answer B (23.5 J) appears to use the initial height of 8.0 m instead of the current height of 3.2 m – this would be the potential energy at the starting position, not the current position. Answer C (14.1 J) might result from incorrectly calculating the distance fallen (8.0 - 3.2 = 4.8 m) and using that as the height, giving $PE = (0.3)(9.8)(4.8) = 14.1 \text{ J}$. Answer D (6.27 J) could stem from using an incorrect value for gravitational acceleration or making an arithmetic error. Remember: gravitational potential energy is always calculated from your reference point upward, regardless of where the object started or how it got to its current position. Focus on the object's present height above your chosen reference level – in this case, the ground.
2. Two identical blocks, each of mass 2.0 kg, are connected by a spring with spring constant 200 N/m. The blocks rest on a frictionless horizontal surface. If one block is displaced 0.15 m from equilibrium and released, what is the maximum potential energy stored in the spring-block system?
   Answer: 2.25 J, occurring at the initial moment of release The maximum potential energy occurs when all the system's energy is potential (when kinetic energy is zero). This happens at the moment of release when the displacement is maximum. $U_{max} = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.15)^2 = \frac{1}{2}(200)(0.0225) = 2.25$ J. Since the surface is frictionless and the system is isolated, this maximum potential energy equals the total mechanical energy. Choice A uses incorrect spring constant. Choice C incorrectly doubles the energy. Choice D uses wrong displacement value or calculation method.
3. Two identical springs with spring constant k are connected to a mass m. In configuration A, the springs are in series (end-to-end). In configuration B, the springs are in parallel (side-by-side). If the mass is displaced by the same distance x from equilibrium in both cases, what is the ratio of potential energy stored in configuration B to that stored in configuration A?
   Answer: 4:1 For springs in series: $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$, so $k_{eq} = \frac{k}{2}$. For springs in parallel: $k_{eq} = k + k = 2k$. The potential energy is $U = \frac{1}{2}k_{eq}x^2$. In configuration A: $U_A = \frac{1}{2} \cdot \frac{k}{2} \cdot x^2 = \frac{kx^2}{4}$. In configuration B: $U_B = \frac{1}{2} \cdot 2k \cdot x^2 = kx^2$. The ratio is $\frac{U_B}{U_A} = \frac{kx^2}{kx^2/4} = 4$. Choice A inverts the ratio. Choice B uses incorrect series combination. Choice C confuses force ratios with energy ratios.
4. A spring with spring constant $k = 400$ N/m is compressed by 0.15 m from its equilibrium position. How much elastic potential energy is stored in the spring?
   Answer: 4.5 J When you encounter spring problems, you're dealing with elastic potential energy, which depends on how much the spring is deformed from its natural length. The key formula here is $U = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from equilibrium. Let's substitute the given values: $k = 400$ N/m and $x = 0.15$ m. The elastic potential energy becomes: $U = \frac{1}{2}(400)(0.15)^2 = \frac{1}{2}(400)(0.0225) = \frac{1}{2}(9) = 4.5 \text{ J}$ This confirms answer A is correct. Now let's examine why the other options are wrong. Answer B (9.0 J) represents what you'd get if you forgot the $\frac{1}{2}$ factor and calculated $kx^2 = 400 \times 0.0225 = 9$ J. Answer C (18 J) occurs if you mistakenly used $kx$ instead of $kx^2$: $400 \times 0.15 \times 0.30 = 18$ J (essentially doubling the displacement). Answer D (30 J) results from the error $kx = 400 \times 0.15 = 60$ J, then somehow halving it, or other calculation mistakes. Remember that elastic potential energy always involves $x^2$, not just $x$, and always includes the $\frac{1}{2}$ factor. This quadratic relationship means small compressions store relatively little energy, while large compressions store dramatically more energy. Double-check your arithmetic with the squared terms, as this is where most calculation errors occur.
5. Two identical masses are connected by a spring on a frictionless horizontal surface. The spring is compressed by 0.10 m from its natural length and then released. When the spring has expanded to 0.05 m beyond its natural length, what fraction of the system's total mechanical energy is stored as elastic potential energy?
   Answer: $\frac{1}{4}$ When analyzing spring-mass systems, you need to apply conservation of mechanical energy and understand how energy transforms between kinetic and potential forms throughout the motion. The total mechanical energy equals the initial elastic potential energy: $E_{total} = \frac{1}{2}k(0.10)^2 = \frac{1}{2}k(0.01) = 0.005k$. This energy remains constant throughout the motion since the surface is frictionless. When the spring extends to 0.05 m beyond its natural length, the elastic potential energy is $U = \frac{1}{2}k(0.05)^2 = \frac{1}{2}k(0.0025) = 0.00125k$. The fraction of total energy stored as elastic potential energy is therefore $\frac{0.00125k}{0.005k} = \frac{1}{4}$, confirming answer A. Let's examine why the other answers are incorrect. Answer B ($\frac{1}{3}$) would require the spring to be compressed or extended by approximately 0.058 m, which doesn't match our 0.05 m extension. Answer C ($\frac{1}{2}$) represents the energy distribution at the equilibrium position in simple harmonic motion, but that's not applicable here since we have two masses connected by a spring. Answer D ($\frac{2}{3}$) would correspond to an extension of about 0.082 m, again not matching our scenario. Remember that elastic potential energy depends on the square of the displacement from equilibrium. When solving these problems, always identify the total energy first (usually from the initial conditions), then calculate the energy at the specified position using $U = \frac{1}{2}kx^2$.
6. Two identical springs, each with spring constant $k = 180$ N/m, are connected in parallel and compressed by 0.20 m. What is the total elastic potential energy stored in the system?
   Answer: 7.2 J When you encounter springs connected in parallel, remember that they share the load and act like a single spring with a combined spring constant. This is a key concept in elastic potential energy problems. In parallel spring systems, the effective spring constant is the sum of individual spring constants: $k_{eff} = k_1 + k_2 = 180 + 180 = 360$ N/m. Both springs compress by the same distance (0.20 m) since they're connected in parallel. The total elastic potential energy is calculated using $U = \frac{1}{2}k_{eff}x^2 = \frac{1}{2}(360)(0.20)^2 = \frac{1}{2}(360)(0.04) = 7.2$ J. This confirms answer A is correct. Looking at the wrong answers: Answer B (3.6 J) results from using only one spring's constant instead of the combined value—this ignores that parallel springs add their spring constants. Answer C (14.4 J) likely comes from incorrectly squaring the compression distance as $(0.20)^2 = 0.04$ but then multiplying by 10 instead of dividing by 2, or from other calculation errors. Answer D (1.8 J) appears to result from using the wrong effective spring constant or mishandling the ½ factor in the energy formula. Study tip: For parallel springs, always add the spring constants first, then treat the system as a single spring. For springs in series, the relationship is $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$. Remember this follows the same pattern as resistors, but inverted—parallel springs add directly, just like parallel resistors have reciprocals that add.
7. Two springs with spring constants $k_1 = 200$ N/m and $k_2 = 300$ N/m are connected in series and compressed by a total distance of 0.20 m from equilibrium. What is the total elastic potential energy stored in the system?
   Answer: 2.4 J When you encounter springs connected in series, remember that they share the same force but have different compressions, and you need to find the equivalent spring constant first. For springs in series, the equivalent spring constant is: $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$ Substituting the values: $\frac{1}{k_{eq}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600}$ Therefore: $k_{eq} = \frac{600}{5} = 120 \text{ N/m}$ The total elastic potential energy is: $U = \frac{1}{2}k_{eq}x^2 = \frac{1}{2}(120)(0.20)^2 = \frac{1}{2}(120)(0.04) = 2.4 \text{ J}$ This confirms answer A) 2.4 J is correct. B) 4.8 J would result if you forgot the factor of ½ in the potential energy formula, giving you $k_{eq}x^2$ instead of $\frac{1}{2}k_{eq}x^2$. C) 10 J comes from incorrectly using the parallel combination formula ($k_{eq} = k_1 + k_2 = 500$ N/m), then calculating $\frac{1}{2}(500)(0.20)^2 = 10$ J. D) 20 J results from both using the wrong parallel formula AND forgetting the ½ factor: $(500)(0.20)^2 = 20$ J. Study tip: Remember that springs in series are "weaker" than either individual spring ($k_{eq} < k_1$ and $k_{eq} < k_2$), while springs in parallel are stronger. Always double-check that your equivalent spring constant makes physical sense before calculating energy.
8. A spring gun fires a 0.05 kg projectile horizontally. The spring has a spring constant of 800 N/m and is compressed 0.15 m before firing. Assuming all elastic potential energy converts to kinetic energy, what was the elastic potential energy stored in the compressed spring?
   Answer: 9.0 J When you encounter spring problems, you're dealing with elastic potential energy, which depends only on the spring constant and compression distance - not the mass of the projectile. The elastic potential energy stored in a compressed spring follows the formula $U = \frac{1}{2}kx^2$, where k is the spring constant and x is the compression distance. Substituting the given values: $U = \frac{1}{2}(800 \text{ N/m})(0.15 \text{ m})^2 = \frac{1}{2}(800)(0.0225) = 9.0 \text{ J}$ This confirms that A) 9.0 J is correct. Let's examine why the other answers are wrong. B) 18 J represents doubling the correct answer - this might result from forgetting the $\frac{1}{2}$ factor in the elastic potential energy formula. C) 4.5 J is exactly half the correct answer, suggesting someone might have incorrectly used $\frac{1}{4}kx^2$ instead of $\frac{1}{2}kx^2$. D) 120 J is far too large and likely comes from incorrectly using the uncompressed spring length or making a significant calculation error with the units. Notice that the projectile's mass (0.05 kg) is irrelevant for calculating stored elastic potential energy - it's only needed if you want to find the projectile's velocity after firing using energy conservation. The key strategy here is recognizing that elastic potential energy depends solely on spring properties (k and x), not on what the spring will eventually accelerate.
9. A 2.0 kg block slides down a frictionless inclined plane from rest. The block starts at a height of 5.0 m above the ground and slides to a height of 1.5 m above the ground. What is the change in gravitational potential energy of the block?
   Answer: -68.6 J (decrease of 68.6 J) When you encounter problems involving objects moving between different heights, you're working with gravitational potential energy, which depends on an object's position in a gravitational field. The key insight is that potential energy change depends only on the initial and final positions, regardless of the path taken. Gravitational potential energy is calculated using $PE = mgh$, where m is mass, g is gravitational acceleration (9.8 m/s²), and h is height. To find the change in potential energy, calculate $\Delta PE = PE_{final} - PE_{initial}$. For this problem: - Initial PE: $PE_i = (2.0 \text{ kg})(9.8 \text{ m/s}^2)(5.0 \text{ m}) = 98 \text{ J}$ - Final PE: $PE_f = (2.0 \text{ kg})(9.8 \text{ m/s}^2)(1.5 \text{ m}) = 29.4 \text{ J}$ - Change: $\Delta PE = 29.4 - 98 = -68.6 \text{ J}$ The negative sign indicates a decrease in potential energy, confirming answer A. Answer B (-29.4 J) incorrectly uses only the final potential energy value as the change. Answer C (+29.4 J) makes the same magnitude error but also gets the wrong sign, suggesting potential energy increased when the block clearly moved to a lower height. Answer D (+68.6 J) has the correct magnitude but wrong sign, perhaps from calculating $PE_{initial} - PE_{final}$ instead of $PE_{final} - PE_{initial}$. Remember: when an object moves to a lower height, gravitational potential energy always decreases (negative change). The "frictionless" detail confirms energy is conserved, converting from potential to kinetic energy.
10. A spring is stretched 0.12 m from its natural length, storing 7.2 J of elastic potential energy. If the same spring is compressed 0.08 m from its natural length, how much elastic potential energy will be stored?
    Answer: 3.2 J When you encounter spring problems, remember that elastic potential energy follows a quadratic relationship with displacement. The formula is $U = \frac{1}{2}kx^2$, where k is the spring constant and x is the displacement from equilibrium. First, you need to find the spring constant using the initial conditions. When stretched 0.12 m with 7.2 J stored: $7.2 = \frac{1}{2}k(0.12)^2$. Solving for k: $k = \frac{2 \times 7.2}{(0.12)^2} = \frac{14.4}{0.0144} = 1000 \text{ N/m}$. Now you can find the energy when compressed 0.08 m: $U = \frac{1}{2}(1000)(0.08)^2 = 500 \times 0.0064 = 3.2 \text{ J}$. This confirms answer A is correct. Looking at the wrong answers: B (4.8 J) would result from incorrectly assuming energy is proportional to displacement rather than displacement squared. C (2.4 J) might come from using the wrong proportionality factor or making arithmetic errors in the calculation. D (1.6 J) could result from forgetting to square the displacement or using an incorrect formula altogether. The key insight is that elastic potential energy depends on the square of displacement, not displacement itself. This means if you double the stretch, you quadruple the energy. Always remember to first find the spring constant from given conditions, then apply it to the new scenario. Watch for the quadratic relationship - it's what makes spring problems different from linear relationships you might expect.
11. A roller coaster car of mass 400 kg starts from rest at the top of a 25 m high hill. Due to friction and air resistance, the car loses 20% of its initial mechanical energy by the time it reaches the bottom. What is the car's kinetic energy at the bottom of the hill?
    Answer: 78,400 J, representing 80% of the initial potential energy Initial mechanical energy equals initial potential energy: $E_i = mgh = (400)(9.8)(25) = 98,000$ J. After losing 20% to friction and air resistance, the remaining energy is: $E_f = 0.80 × 98,000 = 78,400$ J. At the bottom, all remaining mechanical energy is kinetic (potential energy = 0), so $KE = 78,400$ J. Choice B gives the initial mechanical energy without accounting for losses. Choice C incorrectly adds energy. Choice D gives only the energy lost, not the final kinetic energy.
12. A spring with spring constant 320 N/m is first compressed 0.18 m, then released and allowed to extend 0.22 m beyond its natural length. What is the change in elastic potential energy of the spring?
    Answer: +2.56 J When dealing with elastic potential energy problems, you need to calculate the energy at each position and find the difference. The elastic potential energy stored in a spring is given by $U = \frac{1}{2}kx^2$, where k is the spring constant and x is the displacement from the natural length. Let's find the energy at both positions. Initially, the spring is compressed 0.18 m, so $U_1 = \frac{1}{2}(320)(0.18)^2 = \frac{1}{2}(320)(0.0324) = 5.18 \text{ J}$. Finally, the spring is extended 0.22 m beyond its natural length, so $U_2 = \frac{1}{2}(320)(0.22)^2 = \frac{1}{2}(320)(0.0484) = 7.74 \text{ J}$. The change in elastic potential energy is $\Delta U = U_2 - U_1 = 7.74 - 5.18 = +2.56 \text{ J}$. Choice A (+2.56 J) is correct. Choice B (-2.56 J) represents calculating the change backwards (initial minus final instead of final minus initial). Choice C (+5.18 J) is just the initial elastic potential energy when compressed, not the change. Choice D (+7.74 J) is the final elastic potential energy when extended, again not the change between states. Remember that elastic potential energy depends on the square of displacement from equilibrium, regardless of direction. Always calculate the energy at each state first, then find the difference as final minus initial to get the change.
13. A 0.8 kg ball is thrown upward and reaches a maximum height of 12 m above the throwing point. If the ball was thrown from a height of 1.5 m above the ground, what is the gravitational potential energy of the ball at its maximum height, taking ground level as the reference?
    Answer: 106 J This problem tests your understanding of gravitational potential energy and reference points. When calculating gravitational potential energy, you must carefully identify what height to use based on the chosen reference point. Gravitational potential energy is given by $PE = mgh$, where h is the height above the reference point. Since ground level is specified as the reference, you need the total height of the ball above the ground at its maximum point. The ball reaches 12 m above the throwing point, and the throwing point is 1.5 m above ground, so the total height above ground is $h = 12 + 1.5 = 13.5 \text{ m}$. Therefore: $PE = mgh = (0.8 \text{ kg})(9.8 \text{ m/s}^2)(13.5 \text{ m}) = 105.84 \text{ J} \approx 106 \text{ J}$ Answer A (106 J) is correct as it uses the proper total height above the reference point. Answer B (94.1 J) likely uses only the 12 m height above the throwing point, ignoring that ground level is the reference: $(0.8)(9.8)(12) = 94.08 \text{ J}$. Answer C (11.8 J) appears to use only the initial height above ground: $(0.8)(9.8)(1.5) = 11.76 \text{ J}$. Answer D (117 J) might result from calculation errors or using an incorrect value for gravitational acceleration. Always pay close attention to the reference point in potential energy problems. When the reference differs from where motion occurs, you must add or subtract heights accordingly to find the total distance from the reference point.
14. A spring scale is used to weigh a 1.5 kg object by hanging it vertically. The spring stretches 0.25 m from its natural length and the spring constant is 60 N/m. What is the elastic potential energy stored in the stretched spring?
    Answer: 1.875 J When you encounter spring problems involving elastic potential energy, you're dealing with Hooke's law and energy storage in deformed elastic materials. The key relationship here is that elastic potential energy depends on both the spring constant and the square of the displacement. The elastic potential energy stored in a spring is given by $U = \frac{1}{2}kx^2$, where k is the spring constant and x is the displacement from equilibrium. In this problem, you have k = 60 N/m and x = 0.25 m. Substituting these values: $U = \frac{1}{2}(60)(0.25)^2 = \frac{1}{2}(60)(0.0625) = 1.875 \text{ J}$ Answer A (1.875 J) is correct because it properly applies the elastic potential energy formula with the given values. Answer B (3.75 J) represents the common error of forgetting the factor of ½ in the formula. If you incorrectly used $U = kx^2$, you'd get 60 × 0.0625 = 3.75 J. Answer C (15 J) comes from mistakenly using the weight of the object (mg = 1.5 × 10 = 15 N) as if it were the energy, confusing force with energy units. Answer D (0.94 J) might result from calculation errors, possibly mixing up the displacement value or making arithmetic mistakes with the squared term. Remember that elastic potential energy always involves $\frac{1}{2}kx^2$ – the ½ factor is crucial and comes from integrating the variable force over the displacement. Don't confuse the hanging mass with the energy calculation; the mass only determines the equilibrium position, not the energy formula itself.
15. A 3.0 kg object moves from point A at height 4.0 m to point B at height 7.5 m, then to point C at height 2.5 m. What is the change in gravitational potential energy from point A to point C?
    Answer: -44.1 J When you encounter gravitational potential energy problems, focus on the key principle: potential energy depends only on the initial and final positions, not the path taken. The formula is $\Delta PE = mg\Delta h = mg(h_f - h_i)$. To find the change from point A to point C, you only need the starting height (4.0 m) and ending height (2.5 m). Point B is irrelevant—it's just part of the path. Calculate: $\Delta PE = mg(h_C - h_A) = (3.0 \text{ kg})(9.8 \text{ m/s}^2)(2.5 \text{ m} - 4.0 \text{ m}) = (3.0)(9.8)(-1.5) = -44.1 \text{ J}$. The negative sign indicates the object lost potential energy by moving to a lower height. Looking at the wrong answers: Answer B (+44.1 J) has the correct magnitude but wrong sign—this happens if you mistakenly calculate $(h_A - h_C)$ instead of $(h_C - h_A)$. Answer C (-147 J) suggests you incorrectly included point B in your calculation, perhaps finding the total change from A to B to C: $mg(7.5-4.0) + mg(2.5-7.5) = mg(-5.0) = -147 \text{ J}$. Answer D (+147 J) combines both errors: including point B and getting the wrong sign. Remember: gravitational potential energy problems test whether you understand that energy is a state function—only initial and final states matter. Always identify your starting and ending points clearly, and watch your signs carefully when calculating height differences.
16. A 2.5 kg mass is attached to a vertical spring and lowered slowly until the spring is compressed 0.30 m from its natural length. If the spring constant is 250 N/m, what is the elastic potential energy stored in the compressed spring?
    Answer: 11.25 J When you encounter spring problems, you're dealing with elastic potential energy, which depends on how much the spring is compressed or stretched from its natural position. The key formula here is $U = \frac{1}{2}kx^2$, where k is the spring constant and x is the displacement from equilibrium. In this problem, you have all the information you need: k = 250 N/m and x = 0.30 m. Notice that the mass (2.5 kg) is given but isn't needed for calculating elastic potential energy - this is purely about the spring's deformation. Substituting into the formula: $U = \frac{1}{2}(250)(0.30)^2 = \frac{1}{2}(250)(0.09) = 11.25 \text{ J}$ This confirms answer A is correct. Looking at the wrong answers: B (22.5 J) is exactly double the correct answer, which suggests someone forgot the $\frac{1}{2}$ factor and calculated $kx^2$ instead. C (75 J) appears to come from multiplying the mass by gravitational acceleration (2.5 × 10 × 3), confusing elastic potential energy with gravitational potential energy concepts. D (7.35 J) might result from calculation errors or using incorrect values. Remember that elastic potential energy only depends on the spring constant and displacement - the mass affects how much the spring compresses at equilibrium, but once you know the compression distance, mass becomes irrelevant to the energy calculation. Always double-check that you're including the $\frac{1}{2}$ factor in the elastic potential energy formula.
17. A 4.0 kg block slides down a frictionless ramp from rest. The block starts 6.0 m above the bottom of the ramp and slides to a point 1.8 m above the bottom. Taking the bottom of the ramp as the reference level, what is the gravitational potential energy of the block at the final position?
    Answer: 70.6 J Gravitational potential energy problems test your understanding of how energy depends on position in a gravitational field. When you see a block at different heights, focus on the fundamental formula: $PE = mgh$, where m is mass, g is gravitational acceleration, and h is height above your chosen reference level. Here, you need the potential energy at the final position where the block sits 1.8 m above the bottom of the ramp. Since the bottom is your reference level (h = 0), you can directly apply the formula: $PE = mgh = (4.0 \text{ kg})(9.8 \text{ m/s}^2)(1.8 \text{ m}) = 70.6 \text{ J}$ This confirms answer A is correct. Looking at the wrong answers: B (235 J) would result from using the initial height of 6.0 m instead of the final height of 1.8 m – this represents the initial potential energy, not the final. C (164 J) likely comes from calculating the change in potential energy (235 J - 70.6 J ≈ 164 J), but the question asks specifically for the potential energy at the final position, not the change. D (42.3 J) might result from a calculation error, possibly using an incorrect value for g or making an arithmetic mistake. Remember that gravitational potential energy depends only on the object's current height above the reference level – it doesn't matter how the object got there or where it started. Always identify what the question is asking for: initial PE, final PE, or change in PE.
18. A ball is thrown vertically upward from ground level and reaches a maximum height of 20 m before falling back down. If air resistance is negligible, what is the gravitational potential energy of the ball at a height of 8 m above the ground, expressed as a fraction of its maximum potential energy?
    Answer: $\frac{2}{5}$ This problem tests your understanding of gravitational potential energy and how it scales with height. When dealing with potential energy near Earth's surface, remember that gravitational potential energy is directly proportional to height above a reference point. The gravitational potential energy formula is $U = mgh$, where $m$ is mass, $g$ is gravitational acceleration, and $h$ is height. Since we're asked for a fraction comparing two potential energies of the same ball, the mass and gravitational acceleration cancel out, leaving us with a simple ratio of heights. At the maximum height of 20 m, the ball has its maximum potential energy: $U_{max} = mg(20)$. At 8 m height, the potential energy is: $U_8 = mg(8)$. The fraction is therefore: $\frac{U_8}{U_{max}} = \frac{mg(8)}{mg(20)} = \frac{8}{20} = \frac{2}{5}$ Looking at the wrong answers: B) $\frac{1}{3}$ might result from incorrectly using energy conservation equations or confusing kinetic and potential energy relationships. C) $\frac{3}{8}$ could come from inverting parts of the calculation or misapplying the height ratio. D) $\frac{1}{2}$ might stem from incorrectly assuming the midpoint of the trajectory corresponds to half the maximum energy, but 8 m is not halfway to 20 m. The correct answer is A) $\frac{2}{5}$. Study tip: For gravitational potential energy problems near Earth's surface, the energy ratio always equals the height ratio when comparing the same object at different elevations. No complex calculations needed—just set up the fraction of heights.
19. A spring-mass system has a mass attached to a spring with spring constant 150 N/m. When the mass is displaced 0.25 m from equilibrium, the system has 15 J of total mechanical energy. What is the elastic potential energy when the displacement is 0.15 m from equilibrium?
    Answer: 1.69 J When you encounter spring-mass systems with changing displacements, focus on how elastic potential energy varies with position while total mechanical energy remains constant (assuming no friction). First, use the initial conditions to find the amplitude. At 0.25 m displacement with 15 J total energy, you can find the spring's amplitude using energy conservation. The elastic potential energy formula is $U = \frac{1}{2}kx^2$, where k = 150 N/m and x is displacement. At maximum displacement (amplitude A), all energy is potential: $15 = \frac{1}{2}(150)A^2$. However, since we're not at maximum displacement, we need to work with what we know. At x = 0.25 m, if this were maximum displacement, $U = \frac{1}{2}(150)(0.25)^2 = 4.69$ J. But total energy is 15 J, indicating the system has both kinetic and potential energy at this point. For the displacement of 0.15 m: $U = \frac{1}{2}(150)(0.15)^2 = \frac{1}{2}(150)(0.0225) = 1.69$ J. Looking at the wrong answers: B) 5.4 J likely comes from incorrectly using a proportional relationship or miscalculating the spring constant. C) 9.0 J might result from confusing kinetic and potential energy relationships. D) 13.3 J appears to subtract the calculated potential energy from total energy, which would give kinetic energy instead. Remember: elastic potential energy depends only on displacement and spring constant, regardless of the system's total energy. Calculate $U = \frac{1}{2}kx^2$ directly for any given position.
20. An object of mass 1.2 kg is lifted vertically upward at constant velocity from height $h_1 = 2.0$ m to height $h_2 = 5.5$ m above the ground. What is the change in the object's gravitational potential energy?
    Answer: +41.2 J When you encounter problems involving gravitational potential energy, focus on the fundamental relationship: $U = mgh$, where the change in potential energy depends only on the change in height, not the motion details. Since the object moves at constant velocity, the net force is zero, but this doesn't affect the potential energy calculation. The change in gravitational potential energy is $\Delta U = mg(h_2 - h_1)$. Substituting the given values: $\Delta U = (1.2 \text{ kg})(9.8 \text{ m/s}^2)(5.5 - 2.0) \text{ m} = (1.2)(9.8)(3.5) = 41.16 \text{ J}$ Since the object moves to a higher position, the potential energy increases, making this positive: $+41.2 \text{ J}$. Looking at the wrong answers: Choice B gives $-41.2 \text{ J}$, which represents the same magnitude but wrong sign—this would be correct if the object moved downward instead of upward. Choice C shows $+64.7 \text{ J}$, which you'd get if you incorrectly used the final height alone: $mgh_2 = (1.2)(9.8)(5.5) = 64.68 \text{ J}$. Choice D gives $+23.5 \text{ J}$, which could result from calculation errors or using an incorrect value for $g$. The correct answer is A: $+41.2 \text{ J}$. Remember: potential energy problems always depend on the change in height, not absolute position. When an object moves upward, $\Delta U$ is positive; when it moves downward, $\Delta U$ is negative. The motion characteristics (constant velocity, acceleration, etc.) don't directly affect the potential energy calculation.

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