College Physics Quiz: Conservation Of Energy
20 questions · exam conditions
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Conservation Of EnergyQuestion 1 of 20

A ball is thrown vertically upward from ground level with initial speed v0v_0. Air resistance does negative work WrW_r on the ball during its upward flight. In terms of v0v_0, WrW_r, mm, and gg, what is the maximum height reached by the ball?

v022g−Wrmg\frac{v_0^2}{2g} - \frac{W_r}{mg}
v022g+Wrmg\frac{v_0^2}{2g} + \frac{W_r}{mg}
mv02−2Wr2mg\frac{mv_0^2 - 2W_r}{2mg}
mv02+2Wr2mg\frac{mv_0^2 + 2W_r}{2mg}
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College Physics Quiz

College Physics Quiz: Conservation Of Energy

Practice Conservation Of 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 Conservation Of 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

A ball is thrown vertically upward from ground level with initial speed v0v_0. Air resistance does negative work WrW_r on the ball during its upward flight. In terms of v0v_0, WrW_r, mm, and gg, what is the maximum height reached by the ball?

  1. v022g−Wrmg\frac{v_0^2}{2g} - \frac{W_r}{mg} (correct answer)
  2. v022g+Wrmg\frac{v_0^2}{2g} + \frac{W_r}{mg}
  3. mv02−2Wr2mg\frac{mv_0^2 - 2W_r}{2mg}
  4. mv02+2Wr2mg\frac{mv_0^2 + 2W_r}{2mg}
Explanation: Using the work-energy theorem from launch to maximum height: Wnet=ΔK=Kf−Ki=0−12mv02W_{net} = \Delta K = K_f - K_i = 0 - \frac{1}{2}mv_0^2. The net work includes gravity and air resistance: Wnet=Wg+Wr=−mgh+WrW_{net} = W_g + W_r = -mgh + W_r (where WrW_r is negative). So −mgh+Wr=−12mv02-mgh + W_r = -\frac{1}{2}mv_0^2, which gives mgh=12mv02+Wrmgh = \frac{1}{2}mv_0^2 + W_r. Since Wr<0W_r < 0, this becomes h=v022g−∣Wr∣mg=v022g−Wrmgh = \frac{v_0^2}{2g} - \frac{|W_r|}{mg} = \frac{v_0^2}{2g} - \frac{W_r}{mg} (since WrW_r is given as the negative work done). Choice B incorrectly adds the resistance work. Choices C and D have incorrect factors of 2 in the numerator.

Question 2

A pendulum bob of mass 0.50 kg swings from rest at a height of 0.20 m above its lowest point. When the bob passes through the lowest point of its swing, what is its kinetic energy?

  1. 0.49 J
  2. 0.98 J (correct answer)
  3. 1.47 J
  4. 1.96 J
  5. 2.94 J
Explanation: When you encounter pendulum problems, you're dealing with conservation of mechanical energy—the transformation between potential and kinetic energy as the bob swings. At the starting position, the bob has maximum potential energy and zero kinetic energy (since it starts from rest). As it swings down, potential energy converts to kinetic energy. At the lowest point, all the initial potential energy becomes kinetic energy. The initial potential energy is PE=mgh=(0.50 kg)(9.8 m/s2)(0.20 m)=0.98 JPE = mgh = (0.50 \text{ kg})(9.8 \text{ m/s}^2)(0.20 \text{ m}) = 0.98 \text{ J} By conservation of energy, this entire amount becomes kinetic energy at the lowest point, making the answer B) 0.98 J. Looking at the wrong answers: A) 0.49 J represents exactly half the correct value—this might result from incorrectly using g=4.9 m/s2g = 4.9 \text{ m/s}^2 instead of 9.8 m/s29.8 \text{ m/s}^2, or from some other calculation error involving a factor of 2. C) 1.47 J equals 1.5×0.981.5 \times 0.98, suggesting a multiplication error or misapplying a formula. D) 1.96 J is exactly double the correct answer—this could come from incorrectly adding potential and kinetic energies instead of recognizing the conversion, or from using g=19.6 m/s2g = 19.6 \text{ m/s}^2. Remember: In pendulum problems, always identify the reference point for potential energy and apply conservation of mechanical energy. The total mechanical energy remains constant throughout the swing, just shifting between potential and kinetic forms.

Question 3

A roller coaster car of mass 500 kg starts from rest at the top of a 30 m high hill. Assuming no friction, what is the car's speed when it reaches the bottom of the hill?

  1. 12.1 m/s
  2. 17.3 m/s
  3. 24.2 m/s (correct answer)
  4. 29.4 m/s
  5. 34.6 m/s
Explanation: When you encounter a problem involving an object moving under gravity alone, think conservation of energy. This is a classic energy conversion scenario where gravitational potential energy transforms into kinetic energy. At the top of the hill, the roller coaster has maximum potential energy and zero kinetic energy (starts from rest). At the bottom, it has maximum kinetic energy and zero potential energy (taking the bottom as our reference point). Since energy is conserved: PEtop=KEbottomPE_{top} = KE_{bottom} mgh=12mv2mgh = \frac{1}{2}mv^2 Notice the mass cancels out, giving us: gh=12v2gh = \frac{1}{2}v^2 v=2gh=2(9.8)(30)=588=24.2 m/sv = \sqrt{2gh} = \sqrt{2(9.8)(30)} = \sqrt{588} = 24.2 \text{ m/s} Looking at the wrong answers: Choice A (12.1 m/s) represents gh\sqrt{gh}, missing the factor of 2 in the energy equation. Choice B (17.3 m/s) comes from using g=5 m/s2g = 5 \text{ m/s}^2 instead of the correct 9.8 m/s29.8 \text{ m/s}^2. Choice D (29.4 m/s) results from using 2gh2gh instead of 2gh\sqrt{2gh}, forgetting to take the square root. The correct answer is C) 24.2 m/s. Study tip: For energy conservation problems, always identify what type of energy the object has at each point. When mechanical energy is conserved (no friction), simply set initial total energy equal to final total energy. The mass often cancels out, simplifying your calculation significantly.

Question 4

A 1.5 kg ball is thrown vertically upward with an initial speed of 12 m/s. At what height above the launch point will the ball have half of its initial kinetic energy?

  1. 3.7 m (correct answer)
  2. 4.9 m
  3. 6.1 m
  4. 7.3 m
  5. 9.8 m
Explanation: When you encounter projectile motion problems involving energy, think about how kinetic and potential energy transform as the object moves. This question tests your understanding of energy conservation during vertical motion. Initially, the ball has kinetic energy KE0=12mv2=12(1.5)(122)=108 JKE_0 = \frac{1}{2}mv^2 = \frac{1}{2}(1.5)(12^2) = 108 \text{ J}. When the ball reaches a height where it has half this kinetic energy (54 J), you can use energy conservation to find that height. The total mechanical energy remains constant: Etotal=KE0=108 JE_{total} = KE_0 = 108 \text{ J}. At the target height, KE+PE=108 JKE + PE = 108 \text{ J}, where KE=54 JKE = 54 \text{ J} and PE=mghPE = mgh. Therefore: 54+(1.5)(9.8)h=10854 + (1.5)(9.8)h = 108. Solving for h: h=5414.7=3.67 mh = \frac{54}{14.7} = 3.67 \text{ m}, which rounds to 3.7 m (A). Option B (4.9 m) represents a common error where students might calculate the height for one-third of the initial kinetic energy instead of half. Option C (6.1 m) corresponds to calculating the height where kinetic energy equals one-fourth of the initial value. Option D (7.3 m) is close to the maximum height (hmax=v22g=7.35 mh_{max} = \frac{v^2}{2g} = 7.35 \text{ m}), which occurs when all kinetic energy converts to potential energy. Remember: in energy problems, always identify what remains constant (total mechanical energy) and set up your equation accordingly. Double-check whether the question asks for a fraction of initial energy versus remaining energy at a given point.

Question 5

A 0.25 kg block slides across a horizontal surface with an initial speed of 8.0 m/s. Due to friction, it comes to rest after traveling 10 m. How much mechanical energy was lost to friction?

  1. 4.0 J
  2. 8.0 J (correct answer)
  3. 12.0 J
  4. 16.0 J
  5. 20.0 J
Explanation: When you encounter problems involving friction and energy loss, you're dealing with the work-energy theorem and conservation of energy. The key insight is that mechanical energy lost equals the initial kinetic energy when an object comes to rest. The block starts with kinetic energy and ends at rest, so all its initial kinetic energy is converted to heat by friction. Calculate the initial kinetic energy using KE=12mv2KE = \frac{1}{2}mv^2: KE=12(0.25 kg)(8.0 m/s)2=12(0.25)(64)=8.0 JKE = \frac{1}{2}(0.25 \text{ kg})(8.0 \text{ m/s})^2 = \frac{1}{2}(0.25)(64) = 8.0 \text{ J} Since the block comes to complete rest, all 8.0 J of mechanical energy was lost to friction. Looking at the wrong answers: Choice A (4.0 J) might result from forgetting to square the velocity or making an arithmetic error with the kinetic energy formula. Choice C (12.0 J) could come from incorrectly adding the mass and velocity before squaring, or other calculation mistakes. Choice D (16.0 J) likely results from forgetting the 12\frac{1}{2} factor in the kinetic energy formula, giving you just mv2=(0.25)(64)=16mv^2 = (0.25)(64) = 16 J. The correct answer is B) 8.0 J. Strategy tip: In friction problems where objects come to rest, the energy lost always equals the initial kinetic energy. You don't need to calculate the friction force or use the distance traveled—just find 12mv2\frac{1}{2}mv^2. This approach is faster and less error-prone than using work calculations.

Question 6

A 3.0 kg object is lifted vertically upward at constant velocity through a height of 2.0 m in 4.0 s. What is the power required to lift the object?

  1. 7.35 W
  2. 14.7 W (correct answer)
  3. 29.4 W
  4. 58.8 W
  5. 117.6 W
Explanation: When you encounter problems involving constant velocity motion and energy, focus on the relationship between work, energy, and power. Power is the rate at which work is done or energy is transferred. Since the object moves at constant velocity, the net force is zero. This means the applied force exactly balances the gravitational force: F=mg=(3.0 kg)(9.8 m/s2)=29.4 NF = mg = (3.0 \text{ kg})(9.8 \text{ m/s}^2) = 29.4 \text{ N} The work done against gravity is: W=F×d=29.4 N×2.0 m=58.8 JW = F \times d = 29.4 \text{ N} \times 2.0 \text{ m} = 58.8 \text{ J} Power is work divided by time: P=Wt=58.8 J4.0 s=14.7 WP = \frac{W}{t} = \frac{58.8 \text{ J}}{4.0 \text{ s}} = 14.7 \text{ W} This confirms answer B is correct. Answer A (7.35 W) represents half the correct value—you might get this if you mistakenly used g=4.9 m/s2g = 4.9 \text{ m/s}^2 instead of 9.8 m/s29.8 \text{ m/s}^2. Answer C (29.4 W) gives you the applied force value in watts—this happens if you confuse force with power or forget to divide by time. Answer D (58.8 W) is the total work done, expressed in watts—you'd get this if you calculated work correctly but forgot to divide by time to find power. Remember: power problems often involve three steps: find the force needed, calculate the work done, then divide by time. Don't confuse the numerical values of force, work, and power—pay attention to units and what the question is actually asking for.

Question 7

A skier starts from rest at the top of a frictionless slope and reaches the bottom with a speed of 20 m/s. If the skier then encounters a rough horizontal section where the coefficient of kinetic friction is 0.40, how far will the skier slide before coming to rest?

  1. 25.5 m
  2. 51.0 m (correct answer)
  3. 76.5 m
  4. 102 m
  5. 204 m
Explanation: This problem tests your understanding of energy conservation and friction work, two fundamental concepts that often appear together in mechanics problems. The key insight is that the skier's kinetic energy at the bottom of the slope gets completely dissipated by friction work on the horizontal surface. At the bottom, the skier has kinetic energy KE=12mv2=12m(20)2=200mKE = \frac{1}{2}mv^2 = \frac{1}{2}m(20)^2 = 200m joules. On the rough horizontal section, friction does negative work to bring the skier to rest. The friction force is f=μkmg=0.40mgf = \mu_k mg = 0.40mg, so the work done by friction over distance dd is Wf=−μkmgd=−0.40mgdW_f = -\mu_k mgd = -0.40mgd. Setting the initial kinetic energy equal to the work done against friction: 200m=0.40mgd200m = 0.40mgd. The mass cancels out, giving us 200=0.40×9.8×d=3.92d200 = 0.40 \times 9.8 \times d = 3.92d. Solving: d=2003.92=51.0 md = \frac{200}{3.92} = 51.0 \text{ m}. Choice A (25.5 m) represents half the correct distance - likely from forgetting to square the velocity in the kinetic energy formula. Choice C (76.5 m) is 1.5 times the correct answer, possibly from using an incorrect friction coefficient or gravitational value. Choice D (102 m) is exactly double the correct answer, suggesting an error like using 12μk\frac{1}{2}\mu_k instead of μk\mu_k in the friction calculation. Strategy tip: In friction problems, always check that your units work out and remember that kinetic energy depends on v2v^2, not just vv. Setting up energy conservation equations systematically will help you avoid calculation errors.

Question 8

A 0.30 kg object moves in a vertical circle of radius 0.50 m. At the top of the circle, its speed is 2.0 m/s. What is the object's speed at the bottom of the circle?

  1. 3.2 m/s
  2. 4.0 m/s
  3. 4.9 m/s (correct answer)
  4. 5.7 m/s
  5. 6.5 m/s
Explanation: When you encounter a problem involving motion in a vertical circle, think conservation of energy. The object's mechanical energy (kinetic + potential) remains constant throughout its circular path, assuming no friction. Set up your energy conservation equation between the top and bottom of the circle. At the top, the object has both kinetic energy and gravitational potential energy. At the bottom, it has only kinetic energy (taking the bottom as your reference point where PE = 0). Energy at top = Energy at bottom: 12mvtop2+mgh=12mvbottom2\frac{1}{2}mv_{top}^2 + mgh = \frac{1}{2}mv_{bottom}^2 The height difference between top and bottom equals the circle's diameter: h=2r=2(0.50)=1.0 mh = 2r = 2(0.50) = 1.0 \text{ m} Substituting the known values: 12(0.30)(2.0)2+(0.30)(9.8)(1.0)=12(0.30)vbottom2\frac{1}{2}(0.30)(2.0)^2 + (0.30)(9.8)(1.0) = \frac{1}{2}(0.30)v_{bottom}^2 0.60+2.94=0.15vbottom20.60 + 2.94 = 0.15v_{bottom}^2 vbottom=3.540.15=23.6=4.9 m/sv_{bottom} = \sqrt{\frac{3.54}{0.15}} = \sqrt{23.6} = 4.9 \text{ m/s} Option A (3.2 m/s) likely results from forgetting to include potential energy in the calculation. Option B (4.0 m/s) might come from incorrectly using the radius instead of diameter for the height difference. Option D (5.7 m/s) could result from calculation errors or using incorrect values for gravitational acceleration. Remember: In vertical circular motion problems, always use energy conservation and be careful about the height difference—it's the diameter of the circle, not the radius.

Question 9

A 1.0 kg mass is attached to a vertical spring and oscillates with amplitude 0.20 m. If the spring constant is 100 N/m, what is the maximum kinetic energy of the mass during oscillation?

  1. 1.0 J
  2. 2.0 J (correct answer)
  3. 4.0 J
  4. 8.0 J
  5. 10.0 J
Explanation: When you encounter simple harmonic motion problems involving energy, remember that mechanical energy is conserved and transforms between kinetic and potential forms. The maximum kinetic energy occurs when all the system's energy is kinetic (at the equilibrium position), while maximum potential energy occurs at the amplitude points where the mass momentarily stops. For a mass-spring system, the total mechanical energy equals the maximum potential energy: E=12kA2E = \frac{1}{2}kA^2, where k is the spring constant and A is the amplitude. Since energy is conserved, this total energy also equals the maximum kinetic energy. Calculating with the given values: Emax=12(100 N/m)(0.20 m)2=12(100)(0.04)=2.0 JE_{max} = \frac{1}{2}(100 \text{ N/m})(0.20 \text{ m})^2 = \frac{1}{2}(100)(0.04) = 2.0 \text{ J} This confirms answer B is correct. Looking at the wrong answers: A (1.0 J) results from forgetting the factor of 12\frac{1}{2} in the energy formula. C (4.0 J) comes from incorrectly using the amplitude without squaring it: 12(100)(0.20)=10\frac{1}{2}(100)(0.20) = 10, then somehow getting 4.0 J through additional errors. D (8.0 J) might result from using kA2kA^2 without the 12\frac{1}{2} factor: (100)(0.04)=4.0(100)(0.04) = 4.0, then doubling it incorrectly. Study tip: In simple harmonic motion problems, always remember that maximum kinetic energy equals total mechanical energy, which you can find most easily using E=12kA2E = \frac{1}{2}kA^2. Don't forget that crucial factor of 12\frac{1}{2}!

Question 10

A 0.75 kg ball is dropped from a height of 2.5 m onto a concrete floor. If the ball rebounds to a height of 1.8 m, what percentage of the ball's mechanical energy was lost during the collision?

  1. 18%
  2. 28% (correct answer)
  3. 36%
  4. 46%
  5. 72%
Explanation: When you encounter problems involving energy loss during collisions, think about mechanical energy conservation and how to quantify energy dissipation through comparing initial and final states. To find the percentage of mechanical energy lost, you need to compare the ball's initial potential energy with its final potential energy after rebounding. Initially, the ball has gravitational potential energy PEi=mghi=(0.75)(9.8)(2.5)=18.375 JPE_i = mgh_i = (0.75)(9.8)(2.5) = 18.375 \text{ J}. After rebounding to 1.8 m, its potential energy is PEf=mghf=(0.75)(9.8)(1.8)=13.23 JPE_f = mgh_f = (0.75)(9.8)(1.8) = 13.23 \text{ J}. The energy lost is 18.375−13.23=5.145 J18.375 - 13.23 = 5.145 \text{ J}. The percentage lost is 5.14518.375×100%=28%\frac{5.145}{18.375} \times 100\% = 28\%, confirming answer B. Looking at the wrong answers: A (18%) likely comes from incorrectly calculating 1.8−2.52.5×100%\frac{1.8 - 2.5}{2.5} \times 100\%, which treats this as a simple height percentage rather than an energy calculation. C (36%) might result from using the wrong reference point in the percentage calculation, perhaps 5.14513.23×100%\frac{5.145}{13.23} \times 100\%. D (46%) could come from calculation errors or mixing up the energy values. Remember that energy problems often require you to square-check your reference point for percentages. Energy lost should always be calculated as a percentage of the initial energy, not the final energy. Also, since potential energy depends on height linearly (not quadratically like kinetic energy), you can solve these problems using just the height ratio: hi−hfhi\frac{h_i - h_f}{h_i}.

Question 11

A 6.0 kg sledge hammer is lifted 1.2 m above a stake and then dropped. Just before impact, the hammer has a speed of 4.0 m/s. How much energy was lost to air resistance during the fall?

  1. 22.6 J (correct answer)
  2. 48.0 J
  3. 70.6 J
  4. 118.6 J
  5. 166.6 J
Explanation: This problem tests energy conservation and your ability to account for energy losses during motion. When objects fall through air, some mechanical energy is always lost to air resistance, which you can calculate by comparing initial potential energy to final kinetic energy. Start by finding the initial potential energy when the hammer is lifted: PEi=mgh=(6.0 kg)(9.8 m/s2)(1.2 m)=70.6 JPE_i = mgh = (6.0\text{ kg})(9.8\text{ m/s}^2)(1.2\text{ m}) = 70.6\text{ J}. This represents the total mechanical energy available at the start. Next, calculate the kinetic energy just before impact: KEf=12mv2=12(6.0 kg)(4.0 m/s)2=48.0 JKE_f = \frac{1}{2}mv^2 = \frac{1}{2}(6.0\text{ kg})(4.0\text{ m/s})^2 = 48.0\text{ J}. This is how much mechanical energy remains. The energy lost to air resistance equals the difference: 70.6 J−48.0 J=22.6 J70.6\text{ J} - 48.0\text{ J} = 22.6\text{ J}, which is answer A. Looking at the wrong answers: B (48.0 J) is the final kinetic energy, not the energy lost—this confuses what energy remains with what was lost. C (70.6 J) is the initial potential energy, which would only be correct if the hammer had zero speed at impact (impossible unless it stopped mid-air). D (118.6 J) incorrectly adds the potential and kinetic energies rather than finding their difference. Strategy tip: In energy problems involving air resistance, always set up the energy balance: Initial Energy = Final Energy + Energy Lost. The "lost" energy is what you're usually solving for, and it explains why final speeds are often less than what you'd expect from free fall.

Question 12

A 4.0 kg block starts from rest and slides down a frictionless incline of height 3.0 m. At the bottom, it moves horizontally and compresses a spring. If the spring compresses by 0.25 m, what is the average force exerted by the spring during compression?

  1. 235 N
  2. 470 N (correct answer)
  3. 705 N
  4. 940 N
  5. 1410 N
Explanation: This problem combines energy conservation with work-energy relationships, testing your ability to connect motion on an incline with spring compression. Start by using conservation of energy. The block's initial gravitational potential energy converts entirely to kinetic energy at the bottom: PE=mgh=(4.0)(9.8)(3.0)=117.6 JPE = mgh = (4.0)(9.8)(3.0) = 117.6 \text{ J}. This becomes the kinetic energy just before hitting the spring. When the spring compresses, all this kinetic energy converts to elastic potential energy through the work done by the spring force. Using the work-energy theorem: W=Favg⋅d=ΔKE=117.6 JW = F_{avg} \cdot d = \Delta KE = 117.6 \text{ J}. Therefore: Favg=117.60.25=470 NF_{avg} = \frac{117.6}{0.25} = 470 \text{ N}, which is answer B. Looking at the wrong answers: A (235 N) represents exactly half the correct force - this might result from incorrectly using 12kx2\frac{1}{2}kx^2 without properly relating it to the average force. C (705 N) is 1.5 times the correct answer, possibly from confusion about the relationship between maximum and average spring force. D (940 N) is double the correct value, which could come from incorrectly assuming the average force equals the maximum spring force rather than half of it. Remember that for springs, the average force during compression is exactly half the maximum force since spring force varies linearly from zero to maximum. When solving energy problems involving springs, focus on the work-energy relationship: the work done by the average force equals the change in kinetic energy.

Question 13

A spring with spring constant 200 N/m is compressed by 0.30 m. When released, it pushes a 1.0 kg block across a frictionless horizontal surface. What is the maximum speed achieved by the block?

  1. 3.0 m/s
  2. 4.2 m/s (correct answer)
  3. 6.0 m/s
  4. 8.4 m/s
  5. 12.0 m/s
Explanation: This problem tests energy conservation between elastic potential energy and kinetic energy. When you see a compressed spring releasing an object, think about how the stored energy transforms into motion. Initially, all energy is stored as elastic potential energy in the compressed spring: U=12kx2U = \frac{1}{2}kx^2. When the spring releases the block, this energy converts entirely to kinetic energy at maximum speed: K=12mv2K = \frac{1}{2}mv^2. Since energy is conserved: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2. Solving for maximum velocity: v=kx2m=xkmv = \sqrt{\frac{kx^2}{m}} = x\sqrt{\frac{k}{m}} Substituting the values: v=0.302001.0=0.30200=0.30×14.14=4.24 m/sv = 0.30\sqrt{\frac{200}{1.0}} = 0.30\sqrt{200} = 0.30 \times 14.14 = 4.24 \text{ m/s} This rounds to 4.2 m/s, confirming answer B. Answer A (3.0 m/s) likely comes from incorrectly using v=xk/mv = x\sqrt{k/m} but making an arithmetic error or using wrong values. Answer C (6.0 m/s) might result from forgetting the compression distance in the calculation or doubling something incorrectly. Answer D (8.4 m/s) is exactly double the correct answer, suggesting someone might have forgotten the factor of 12\frac{1}{2} in the energy equations or made a similar systematic error. Remember: in spring-mass problems, maximum speed occurs when all potential energy converts to kinetic energy. Set Uspring=KmaxU_{spring} = K_{max} and solve directly for velocity using energy conservation.

Question 14

A 0.50 kg ball is attached to a string and swings as a pendulum. At the lowest point of its swing, the ball has a speed of 3.0 m/s. What is the maximum height above the lowest point that the ball will reach?

  1. 0.23 m
  2. 0.46 m (correct answer)
  3. 0.69 m
  4. 0.92 m
  5. 1.38 m
Explanation: This is a classic energy conservation problem involving a pendulum. When you see a pendulum question asking about maximum height, think about how kinetic energy converts to gravitational potential energy. At the lowest point, the ball has maximum kinetic energy (KE=12mv2KE = \frac{1}{2}mv^2) and minimum potential energy (we set this as our reference point, so PE=0PE = 0). At the highest point, all kinetic energy converts to gravitational potential energy (PE=mghPE = mgh), so the ball momentarily stops before swinging back. Using conservation of energy: KEbottom=PEtopKE_{bottom} = PE_{top} 12mv2=mgh\frac{1}{2}mv^2 = mgh The mass cancels out: 12v2=gh\frac{1}{2}v^2 = gh Solving for height: h=v22g=(3.0)22(9.8)=9.019.6=0.46 mh = \frac{v^2}{2g} = \frac{(3.0)^2}{2(9.8)} = \frac{9.0}{19.6} = 0.46 \text{ m} Looking at the wrong answers: (A) 0.23 m results from forgetting the factor of 2 in the kinetic energy formula, calculating h=v24gh = \frac{v^2}{4g}. (C) 0.69 m comes from using g=10g = 10 m/s² but making an arithmetic error, or incorrectly applying h=3v22gh = \frac{3v^2}{2g}. (D) 0.92 m represents roughly doubling the correct answer, perhaps from incorrectly setting up the energy equation. The correct answer is (B) 0.46 m. Study tip: In pendulum problems, the mass always cancels out when using energy conservation. Focus on the energy conversion: all kinetic energy at the bottom becomes potential energy at the top.

Question 15

A 1.2 kg book falls from rest from a height of 1.8 m onto a table that is 0.8 m high. What is the book's kinetic energy just before it hits the table?

  1. 9.4 J
  2. 11.8 J (correct answer)
  3. 18.8 J
  4. 21.2 J
  5. 30.6 J
Explanation: When you encounter problems involving falling objects, you're dealing with energy conservation. The key insight is that gravitational potential energy converts to kinetic energy as an object falls. The book falls from a height of 1.8 m to a table at 0.8 m, so it actually falls through a distance of 1.8−0.8=1.0 m1.8 - 0.8 = 1.0 \text{ m}. This is the crucial step many students miss—you need the actual distance fallen, not the initial height. Using conservation of energy, the gravitational potential energy lost equals the kinetic energy gained: KE=mgh=(1.2 kg)(9.8 m/s2)(1.0 m)=11.8 JKE = mgh = (1.2 \text{ kg})(9.8 \text{ m/s}^2)(1.0 \text{ m}) = 11.8 \text{ J}. This confirms answer B is correct. Let's examine why the other answers are wrong. Answer A (9.4 J) appears to use an incorrect value for gravitational acceleration, possibly 7.8 m/s² instead of 9.8 m/s². Answer C (18.8 J) results from using the full initial height of 1.8 m instead of the actual fall distance of 1.0 m—this is the most common error on problems like this. Answer D (21.2 J) comes from incorrectly using the total height from ground to initial position (1.8 + 0.8 = 2.6 m), showing a fundamental misunderstanding of the setup. Always identify the actual distance through which the object falls by finding the difference between initial and final heights. Don't let the presence of intermediate surfaces (like the table) distract you from calculating the correct displacement.

Question 16

A compressed spring with spring constant 150 N/m stores 12 J of elastic potential energy. By how much is the spring compressed?

  1. 0.20 m
  2. 0.28 m
  3. 0.40 m (correct answer)
  4. 0.57 m
  5. 0.80 m
Explanation: When you encounter spring problems involving stored energy, you're working with elastic potential energy, which follows the relationship U=12kx2U = \frac{1}{2}kx^2, where U is the stored energy, k is the spring constant, and x is the compression or extension distance. Given that the spring stores 12 J of energy with a spring constant of 150 N/m, you can solve for the compression distance. Rearranging the elastic potential energy formula: x=2Uk=2(12)150=24150=0.16=0.40 mx = \sqrt{\frac{2U}{k}} = \sqrt{\frac{2(12)}{150}} = \sqrt{\frac{24}{150}} = \sqrt{0.16} = 0.40 \text{ m} Looking at the wrong answers: Choice A (0.20 m) represents a common error where students might forget to take the square root, calculating 2Uk=0.16\frac{2U}{k} = 0.16 and mistakenly using 0.04=0.20\sqrt{0.04} = 0.20. Choice B (0.28 m) could result from computational errors in the square root calculation or mixing up the formula. Choice D (0.57 m) might come from using an incorrect form of the energy equation, such as U=kx2U = kx^2 instead of U=12kx2U = \frac{1}{2}kx^2, which would give x=12150≈0.28x = \sqrt{\frac{12}{150}} ≈ 0.28, then making additional calculation errors. The correct answer is C (0.40 m). Remember: elastic potential energy problems always involve that factor of 12\frac{1}{2} in the formula. Write down U=12kx2U = \frac{1}{2}kx^2 first, then carefully solve for the unknown variable. Double-check your square root calculations, as they're a frequent source of errors.

Question 17

A spring-powered toy car has a spring compressed by 0.15 m. When released, the 0.20 kg car reaches a maximum speed of 1.8 m/s on a horizontal surface. What is the spring constant?

  1. 14.4 N/m
  2. 28.8 N/m (correct answer)
  3. 57.6 N/m
  4. 72.0 N/m
  5. 144 N/m
Explanation: When you encounter a spring-powered system, you're dealing with energy conservation. The elastic potential energy stored in the compressed spring converts entirely to kinetic energy when the car reaches maximum speed (assuming no friction losses). Start with the energy conservation equation: PEspring=KEmaxPE_{spring} = KE_{max} The elastic potential energy formula is PE=12kx2PE = \frac{1}{2}kx^2, where k is the spring constant and x is compression distance. The kinetic energy formula is KE=12mv2KE = \frac{1}{2}mv^2. Setting them equal: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2 Solving for k: k=mv2x2k = \frac{mv^2}{x^2} Substituting the given values: k=(0.20 kg)(1.8 m/s)2(0.15 m)2=(0.20)(3.24)0.0225=28.8 N/mk = \frac{(0.20\text{ kg})(1.8\text{ m/s})^2}{(0.15\text{ m})^2} = \frac{(0.20)(3.24)}{0.0225} = 28.8\text{ N/m} This confirms answer B is correct. Answer A (14.4 N/m) results from forgetting to square the velocity in the calculation. Answer C (57.6 N/m) comes from doubling the correct answer, possibly by incorrectly keeping both 12\frac{1}{2} factors in the energy equation. Answer D (72.0 N/m) occurs when you forget to square the compression distance, using just x instead of x². Study tip: In energy conservation problems, carefully track which quantities get squared in the formulas. Both kinetic energy and elastic potential energy involve squared terms (v² and x²), and forgetting these squares is the most common calculation error on physics exams.

Question 18

A 2.0 kg block slides down a frictionless incline from height h1=3.0h_1 = 3.0 m, then moves across a rough horizontal surface with coefficient of kinetic friction μk=0.25\mu_k = 0.25, and finally compresses a spring with spring constant k=400k = 400 N/m by a maximum distance of 0.50 m. What was the distance the block traveled across the rough horizontal surface?

  1. 4.0 m
  2. 4.9 m
  3. 5.4 m (correct answer)
  4. 6.1 m
Explanation: Using conservation of energy: Initial gravitational PE = Final elastic PE + Work done against friction. mgh1=12kx2+μkmgdmgh_1 = \frac{1}{2}kx^2 + \mu_k mg d. Substituting values: (2.0)(9.8)(3.0)=12(400)(0.50)2+(0.25)(2.0)(9.8)d(2.0)(9.8)(3.0) = \frac{1}{2}(400)(0.50)^2 + (0.25)(2.0)(9.8)d. This gives 58.8=50+4.9d58.8 = 50 + 4.9d, so d=8.8/1.6=5.4d = 8.8/1.6 = 5.4 m. Choice A neglects the spring's stored energy. Choice B uses incorrect friction calculation. Choice D incorrectly adds energies instead of balancing them.

Question 19

A block slides down a frictionless incline of angle θ\theta and height hh, then moves across a horizontal surface with kinetic friction coefficient μk\mu_k before coming to rest. If the incline angle is doubled while keeping the height constant, how does the distance traveled on the horizontal surface change?

  1. The distance decreases because the block spends less time on the incline
  2. The distance increases because the block has greater speed at the bottom
  3. The distance decreases because the normal force component changes on the incline
  4. The distance remains the same because the initial potential energy is unchanged (correct answer)
Explanation: This problem tests energy conservation and how different path geometries affect motion outcomes. When analyzing multi-stage motion problems, focus on what quantities are conserved throughout the entire process. The key insight is that energy conservation governs the entire motion. Initially, the block has gravitational potential energy mghmgh. As it slides down the frictionless incline, this converts completely to kinetic energy 12mv2\frac{1}{2}mv^2 at the bottom, regardless of the incline angle. The speed at the bottom depends only on the height: v=2ghv = \sqrt{2gh}. On the horizontal surface, friction does negative work until the block stops. The work-energy theorem tells us that the kinetic energy at the bottom equals the work done by friction: 12mv2=μkmg⋅d\frac{1}{2}mv^2 = \mu_k mg \cdot d, where dd is the stopping distance. Since v=2ghv = \sqrt{2gh}, we get d=hμkd = \frac{h}{\mu_k}. This distance depends only on the height and friction coefficient, not the incline angle. Choice A incorrectly focuses on time spent on the incline, which doesn't affect the final energy. Choice B wrongly assumes that doubling the angle increases the bottom speed, but speed depends only on height, not angle. Choice C mentions normal force changes on the incline, but since the incline is frictionless, normal forces there don't affect the motion's outcome. Study tip: In energy problems involving multiple surfaces, always trace energy transformations from start to finish. Conservative forces (like gravity) and the total energy budget determine outcomes, while path details often don't matter.

Question 20

A particle moves in a conservative force field where the potential energy function is U(x)=ax2−bx4U(x) = ax^2 - bx^4, where aa and bb are positive constants. If the particle has total mechanical energy EE and is moving with speed vv at position xx, which expression correctly represents the conservation of energy?

  1. E=12mv2−ax2+bx4E = \frac{1}{2}mv^2 - ax^2 + bx^4
  2. E=12mv2+ax2−bx4E = \frac{1}{2}mv^2 + ax^2 - bx^4 (correct answer)
  3. E=12mv2+(2ax−4bx3)xE = \frac{1}{2}mv^2 + (2ax - 4bx^3)x
  4. E=12mv2−(2ax−4bx3)xE = \frac{1}{2}mv^2 - (2ax - 4bx^3)x
Explanation: When you encounter conservative force problems, remember that mechanical energy is always conserved, meaning the sum of kinetic and potential energy remains constant throughout the motion. The conservation of mechanical energy states that E=K+UE = K + U, where KK is kinetic energy and UU is potential energy. Since kinetic energy is K=12mv2K = \frac{1}{2}mv^2 and the given potential energy function is U(x)=ax2−bx4U(x) = ax^2 - bx^4, the total mechanical energy becomes E=12mv2+ax2−bx4E = \frac{1}{2}mv^2 + ax^2 - bx^4. This matches option B exactly. Option A incorrectly uses E=12mv2−ax2+bx4E = \frac{1}{2}mv^2 - ax^2 + bx^4, which suggests the potential energy is −U(x)-U(x). This represents a common sign error where students mistakenly think potential energy should be subtracted from kinetic energy, but energy conservation requires adding them. Options C and D both use the expression (2ax−4bx3)x(2ax - 4bx^3)x, which equals 2ax2−4bx42ax^2 - 4bx^4. These students likely confused potential energy with force. The force is F=−dUdx=−(2ax−4bx3)F = -\frac{dU}{dx} = -(2ax - 4bx^3), so they mistakenly used force-related expressions instead of the potential energy function itself. Option C adds this incorrect term while option D subtracts it. Remember: mechanical energy conservation always uses E=K+UE = K + U directly. Don't overthink it by involving forces or derivatives—just add the kinetic energy to the given potential energy function as written.