College Physics Quiz: Potential Energy
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Potential EnergyQuestion 1 of 20

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?

14\frac{1}{4}
13\frac{1}{3}
12\frac{1}{2}
23\frac{2}{3}
34\frac{3}{4}
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College Physics Quiz

College Physics Quiz: Potential Energy

Practice Potential Energy in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Potential Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

All questions

Question 1

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?

  1. 14\frac{1}{4} (correct answer)
  2. 13\frac{1}{3}
  3. 12\frac{1}{2}
  4. 23\frac{2}{3}
  5. 34\frac{3}{4}
Explanation: 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: Etotal=12k(0.10)2=12k(0.01)=0.005kE_{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=12k(0.05)2=12k(0.0025)=0.00125kU = \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 0.00125k0.005k=14\frac{0.00125k}{0.005k} = \frac{1}{4}, confirming answer A. Let's examine why the other answers are incorrect. Answer B (13\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 (12\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 (23\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=12kx2U = \frac{1}{2}kx^2.

Question 2

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?

  1. 25\frac{2}{5} (correct answer)
  2. 13\frac{1}{3}
  3. 38\frac{3}{8}
  4. 12\frac{1}{2}
  5. 35\frac{3}{5}
Explanation: 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=mghU = mgh, where mm is mass, gg is gravitational acceleration, and hh 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: Umax=mg(20)U_{max} = mg(20). At 8 m height, the potential energy is: U8=mg(8)U_8 = mg(8). The fraction is therefore: U8Umax=mg(8)mg(20)=820=25\frac{U_8}{U_{max}} = \frac{mg(8)}{mg(20)} = \frac{8}{20} = \frac{2}{5} Looking at the wrong answers: B) 13\frac{1}{3} might result from incorrectly using energy conservation equations or confusing kinetic and potential energy relationships. C) 38\frac{3}{8} could come from inverting parts of the calculation or misapplying the height ratio. D) 12\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) 25\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.

Question 3

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?

  1. -68.6 J (decrease of 68.6 J) (correct answer)
  2. -29.4 J (decrease of 29.4 J)
  3. +29.4 J (increase of 29.4 J)
  4. +68.6 J (increase of 68.6 J)
  5. -98.0 J (decrease of 98.0 J)
Explanation: 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=mghPE = 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 ΔPE=PEfinal−PEinitial\Delta PE = PE_{final} - PE_{initial}. For this problem:
  • Initial PE: PEi=(2.0 kg)(9.8 m/s2)(5.0 m)=98 JPE_i = (2.0 \text{ kg})(9.8 \text{ m/s}^2)(5.0 \text{ m}) = 98 \text{ J}
  • Final PE: PEf=(2.0 kg)(9.8 m/s2)(1.5 m)=29.4 JPE_f = (2.0 \text{ kg})(9.8 \text{ m/s}^2)(1.5 \text{ m}) = 29.4 \text{ J}
  • Change: ΔPE=29.4−98=−68.6 J\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 PEinitial−PEfinalPE_{initial} - PE_{final} instead of PEfinal−PEinitialPE_{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.

Question 4

A spring with spring constant k=400k = 400 N/m is compressed by 0.15 m from its equilibrium position. How much elastic potential energy is stored in the spring?

  1. 4.5 J (correct answer)
  2. 9.0 J
  3. 18 J
  4. 30 J
  5. 60 J
Explanation: 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=12kx2U = \frac{1}{2}kx^2, where kk is the spring constant and xx is the displacement from equilibrium. Let's substitute the given values: k=400k = 400 N/m and x=0.15x = 0.15 m. The elastic potential energy becomes: U=12(400)(0.15)2=12(400)(0.0225)=12(9)=4.5 JU = \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 12\frac{1}{2} factor and calculated kx2=400×0.0225=9kx^2 = 400 \times 0.0225 = 9 J. Answer C (18 J) occurs if you mistakenly used kxkx instead of kx2kx^2: 400×0.15×0.30=18400 \times 0.15 \times 0.30 = 18 J (essentially doubling the displacement). Answer D (30 J) results from the error kx=400×0.15=60kx = 400 \times 0.15 = 60 J, then somehow halving it, or other calculation mistakes. Remember that elastic potential energy always involves x2x^2, not just xx, and always includes the 12\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.

Question 5

A roller coaster car of mass 500 kg is at the top of a hill 30 m high. Taking the bottom of the hill as the reference level for gravitational potential energy, what is the gravitational potential energy of the car at the top?

  1. 147,000 J (correct answer)
  2. 73,500 J
  3. 98,000 J
  4. 196,000 J
  5. 294,000 J
Explanation: When you encounter gravitational potential energy problems, you're working with the fundamental relationship between an object's position in a gravitational field and its stored energy. The key formula is PE=mghPE = mgh, where m is mass, g is gravitational acceleration (9.8 m/s²), and h is height above the reference level. Here, you have a 500 kg roller coaster car at 30 m height, with the bottom of the hill as your reference point (h = 0). Substituting into the formula: PE=(500 kg)(9.8 m/s2)(30 m)=147,000 JPE = (500 \text{ kg})(9.8 \text{ m/s}^2)(30 \text{ m}) = 147,000 \text{ J} Looking at the incorrect choices: Answer B (73,500 J) represents exactly half the correct value, suggesting someone might have used 4.9 m/s² instead of 9.8 m/s² for gravity—a common error from confusing gravitational acceleration with other constants. Answer C (98,000 J) appears to come from using g ≈ 6.5 m/s², which has no physical basis. Answer D (196,000 J) is roughly 4/3 times the correct answer, possibly from computational errors or using an incorrect value around 13 m/s² for gravity. The correct answer is A: 147,000 J. Remember that gravitational potential energy problems are straightforward applications of PE=mghPE = mgh. Always double-check that you're using g = 9.8 m/s² (or 10 m/s² if specified), and make sure your height measurement is from the stated reference level. These problems test your ability to apply the formula correctly rather than conceptual complexity.

Question 6

A pendulum bob of mass 0.5 kg swings from its lowest point to a position where it is 0.8 m higher than the lowest point. If we take the lowest point as our reference level, what is the gravitational potential energy at the higher position?

  1. 3.92 J (correct answer)
  2. 1.96 J
  3. 7.84 J
  4. 0.4 J
  5. 4.9 J
Explanation: This question tests gravitational potential energy, a fundamental concept in energy conservation problems. When you see a pendulum or any object moving vertically, think about how height changes affect potential energy. Gravitational potential energy is calculated using the formula U=mghU = mgh, where mm is mass, gg is gravitational acceleration (9.8 m/s²), and hh is height above the reference point. Since the problem states the lowest point is our reference level, and the bob rises 0.8 m above this point, we can substitute directly: U=(0.5 kg)(9.8 m/s2)(0.8 m)=3.92 JU = (0.5 \text{ kg})(9.8 \text{ m/s}^2)(0.8 \text{ m}) = 3.92 \text{ J} This confirms answer A is correct. Looking at the wrong answers: Answer B (1.96 J) represents exactly half the correct value, suggesting someone might have used g=4.9 m/s2g = 4.9 \text{ m/s}^2 instead of 9.8, or made an arithmetic error. Answer C (7.84 J) is double the correct answer, possibly from using the wrong mass or height value. Answer D (0.4 J) appears to come from multiplying just the mass and height (0.5 × 0.8) while forgetting to include gravitational acceleration entirely. Remember that gravitational potential energy problems always require all three variables: mass, gravitational acceleration, and height. The most common mistake is forgetting to include g=9.8 m/s2g = 9.8 \text{ m/s}^2 in your calculation. Always double-check that you've used the complete formula U=mghU = mgh.

Question 7

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?

  1. 1.69 J (correct answer)
  2. 5.4 J
  3. 9.0 J
  4. 13.3 J
  5. 2.25 J
Explanation: 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=12kx2U = \frac{1}{2}kx^2, where k = 150 N/m and x is displacement. At maximum displacement (amplitude A), all energy is potential: 15=12(150)A215 = \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=12(150)(0.25)2=4.69U = \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=12(150)(0.15)2=12(150)(0.0225)=1.69U = \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=12kx2U = \frac{1}{2}kx^2 directly for any given position.

Question 8

Two identical springs, each with spring constant k=180k = 180 N/m, are connected in parallel and compressed by 0.20 m. What is the total elastic potential energy stored in the system?

  1. 7.2 J (correct answer)
  2. 3.6 J
  3. 14.4 J
  4. 1.8 J
  5. 36 J
Explanation: 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: keff=k1+k2=180+180=360k_{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=12keffx2=12(360)(0.20)2=12(360)(0.04)=7.2U = \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(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 1keff=1k1+1k2\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.

Question 9

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?

  1. 106 J (correct answer)
  2. 94.1 J
  3. 11.8 J
  4. 117 J
  5. 82.3 J
Explanation: 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=mghPE = 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 mh = 12 + 1.5 = 13.5 \text{ m}. Therefore: PE=mgh=(0.8 kg)(9.8 m/s2)(13.5 m)=105.84 J≈106 JPE = 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 J(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 J(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.

Question 10

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?

  1. 9.0 J (correct answer)
  2. 18 J
  3. 4.5 J
  4. 120 J
  5. 60 J
Explanation: 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=12kx2U = \frac{1}{2}kx^2, where k is the spring constant and x is the compression distance. Substituting the given values: U=12(800 N/m)(0.15 m)2=12(800)(0.0225)=9.0 JU = \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 12\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 14kx2\frac{1}{4}kx^2 instead of 12kx2\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.

Question 11

An object of mass 1.2 kg is lifted vertically upward at constant velocity from height h1=2.0h_1 = 2.0 m to height h2=5.5h_2 = 5.5 m above the ground. What is the change in the object's gravitational potential energy?

  1. +41.2 J (correct answer)
  2. -41.2 J
  3. +64.7 J
  4. +23.5 J
  5. +129.4 J
Explanation: When you encounter problems involving gravitational potential energy, focus on the fundamental relationship: U=mghU = 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 ΔU=mg(h2−h1)\Delta U = mg(h_2 - h_1). Substituting the given values: ΔU=(1.2 kg)(9.8 m/s2)(5.5−2.0) m=(1.2)(9.8)(3.5)=41.16 J\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 J+41.2 \text{ J}. Looking at the wrong answers: Choice B gives −41.2 J-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 J+64.7 \text{ J}, which you'd get if you incorrectly used the final height alone: mgh2=(1.2)(9.8)(5.5)=64.68 Jmgh_2 = (1.2)(9.8)(5.5) = 64.68 \text{ J}. Choice D gives +23.5 J+23.5 \text{ J}, which could result from calculation errors or using an incorrect value for gg. The correct answer is A: +41.2 J+41.2 \text{ J}. Remember: potential energy problems always depend on the change in height, not absolute position. When an object moves upward, ΔU\Delta U is positive; when it moves downward, ΔU\Delta U is negative. The motion characteristics (constant velocity, acceleration, etc.) don't directly affect the potential energy calculation.

Question 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?

  1. +2.56 J (correct answer)
  2. -2.56 J
  3. +5.18 J
  4. +7.74 J
  5. 0 J
Explanation: 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=12kx2U = \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 U1=12(320)(0.18)2=12(320)(0.0324)=5.18 JU_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 U2=12(320)(0.22)2=12(320)(0.0484)=7.74 JU_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 ΔU=U2−U1=7.74−5.18=+2.56 J\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.

Question 13

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?

  1. 70.6 J (correct answer)
  2. 235 J
  3. 164 J
  4. 42.3 J
  5. 94.1 J
Explanation: 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=mghPE = 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 kg)(9.8 m/s2)(1.8 m)=70.6 JPE = 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.

Question 14

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?

  1. 9.41 J (correct answer)
  2. 23.5 J
  3. 14.1 J
  4. 6.27 J
  5. 31.4 J
Explanation: 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=mghPE = 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 kg)(9.8 m/s2)(3.2 m)=9.41 JPE = (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 JPE = (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.

Question 15

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?

  1. 1.875 J (correct answer)
  2. 3.75 J
  3. 15 J
  4. 0.94 J
  5. 7.5 J
Explanation: 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=12kx2U = \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=12(60)(0.25)2=12(60)(0.0625)=1.875 JU = \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=kx2U = 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 12kx2\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.

Question 16

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?

  1. -44.1 J (correct answer)
  2. +44.1 J
  3. -147 J
  4. +147 J
  5. 0 J
Explanation: 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 ΔPE=mgΔh=mg(hf−hi)\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: ΔPE=mg(hC−hA)=(3.0 kg)(9.8 m/s2)(2.5 m−4.0 m)=(3.0)(9.8)(−1.5)=−44.1 J\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 (hA−hC)(h_A - h_C) instead of (hC−hA)(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 Jmg(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.

Question 17

Two springs with spring constants k1=200k_1 = 200 N/m and k2=300k_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?

  1. 2.4 J (correct answer)
  2. 4.8 J
  3. 10 J
  4. 20 J
  5. 50 J
Explanation: 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: 1keq=1k1+1k2\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} Substituting the values: 1keq=1200+1300=3+2600=5600\frac{1}{k_{eq}} = \frac{1}{200} + \frac{1}{300} = \frac{3 + 2}{600} = \frac{5}{600} Therefore: keq=6005=120 N/mk_{eq} = \frac{600}{5} = 120 \text{ N/m} The total elastic potential energy is: U=12keqx2=12(120)(0.20)2=12(120)(0.04)=2.4 JU = \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 keqx2k_{eq}x^2 instead of 12keqx2\frac{1}{2}k_{eq}x^2. C) 10 J comes from incorrectly using the parallel combination formula (keq=k1+k2=500k_{eq} = k_1 + k_2 = 500 N/m), then calculating 12(500)(0.20)2=10\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(500)(0.20)^2 = 20 J. Study tip: Remember that springs in series are "weaker" than either individual spring (keq<k1k_{eq} < k_1 and keq<k2k_{eq} < k_2), while springs in parallel are stronger. Always double-check that your equivalent spring constant makes physical sense before calculating energy.

Question 18

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?

  1. 11.25 J (correct answer)
  2. 22.5 J
  3. 75 J
  4. 7.35 J
  5. 18.4 J
Explanation: 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=12kx2U = \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=12(250)(0.30)2=12(250)(0.09)=11.25 JU = \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 12\frac{1}{2} factor and calculated kx2kx^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 12\frac{1}{2} factor in the elastic potential energy formula.

Question 19

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?

  1. 3.2 J (correct answer)
  2. 4.8 J
  3. 2.4 J
  4. 1.6 J
  5. 6.4 J
Explanation: When you encounter spring problems, remember that elastic potential energy follows a quadratic relationship with displacement. The formula is U=12kx2U = \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=12k(0.12)27.2 = \frac{1}{2}k(0.12)^2. Solving for k: k=2×7.2(0.12)2=14.40.0144=1000 N/mk = \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=12(1000)(0.08)2=500×0.0064=3.2 JU = \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.

Question 20

A 1.5 kg book is lifted from the floor to a shelf 2.2 m high, then moved horizontally 1.8 m along the shelf to its final position. Taking the floor as the reference level, what is the gravitational potential energy of the book in its final position?

  1. 32.3 J (correct answer)
  2. 26.5 J
  3. 58.8 J
  4. 16.2 J
  5. 48.6 J
Explanation: When you encounter gravitational potential energy problems, focus on the vertical displacement only—horizontal movement doesn't affect gravitational potential energy since gravity acts vertically downward. Gravitational potential energy is calculated using U=mghU = mgh, where mm is mass, gg is gravitational acceleration (9.8 m/s²), and hh is the height above the reference level. Here, the book's final height is 2.2 m above the floor (our reference level), regardless of its horizontal position along the shelf. U=(1.5 kg)(9.8 m/s2)(2.2 m)=32.34 JU = (1.5 \text{ kg})(9.8 \text{ m/s}^2)(2.2 \text{ m}) = 32.34 \text{ J} This rounds to 32.3 J, confirming answer A is correct. Looking at the wrong answers: B (26.5 J) might result from using an incorrect value for gg or making an arithmetic error. C (58.8 J) is nearly double the correct answer—this could happen if you mistakenly tried to include the horizontal displacement in your calculation, perhaps adding the vertical and horizontal distances (2.2 + 1.8 = 4.0 m) before calculating. D (16.2 J) is exactly half the correct answer, suggesting a calculation error like using 4.9 m/s² instead of 9.8 m/s² for gravity. Remember: gravitational potential energy depends only on vertical position relative to your reference level. Horizontal motion at constant height doesn't change gravitational potential energy—only the work done against gravity (vertical displacement) matters. Always identify your reference level first, then focus solely on the vertical height difference.