AP Physics 1 Quiz: Conservation Of Energy

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 AP Physics 1.

How to use this quiz

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

Question 1

1. $v=\sqrt{2gh}$
2. $v=\sqrt{\dfrac{2(mgh-E_d)}{m}}$ (correct answer)
3. $v=\sqrt{\dfrac{2(mgh+E_d)}{m}}$
4. $v=\sqrt{2gh}-\sqrt{\dfrac{2E_d}{m}}$

Explanation: This problem involves energy conservation with energy dissipation by air drag. The energy equation is: initial gravitational potential energy equals final kinetic energy plus dissipated energy, or $mgh = \frac{1}{2}mv^2 + E_d$. Rearranging to solve for $v$: $mgh - E_d = \frac{1}{2}mv^2$, so $v^2 = \frac{2(mgh - E_d)}{m}$, and therefore $v = \sqrt{\frac{2(mgh - E_d)}{m}}$. Choice C incorrectly adds $E_d$ to the initial energy, which would mean air drag adds energy rather than removing it. When dealing with energy dissipation, always subtract the dissipated energy from the initial mechanical energy to find the remaining energy.

Question 2

1. $v=\sqrt{v_0^2+2gH}$ (correct answer)
2. $v=v_0+\sqrt{2gH}$
3. $v=\sqrt{v_0^2-2gH}$
4. $v=\sqrt{2gH-v_0^2}$

Explanation: This question applies conservation of energy to a ball falling after rolling off a table. The initial kinetic energy is $(\frac{1}{2}) m v_0^2$ (horizontal), and potential energy decreases by $m g H$. The vertical speed component is $\sqrt{2 g H}$, so total v = $\sqrt{v_0^2 + 2 g H}$. Energy is conserved separately in horizontal and vertical directions with no air resistance. Option B is incorrect as it adds speeds linearly instead of using vector magnitude. In combined motion problems, separate components and combine using Pythagoras for total speed before impact.

Question 3

1. $0$
2. $+mgh$
3. $-mgh$ (correct answer)
4. $-\vec g\cdot \vec h$

Explanation: This problem involves choosing a reference point for gravitational potential energy and calculating potential energy at a different location. The student sets $U_g = 0$ at the release point (height $h$ above the table), which means the potential energy reference is at height $h$. At the table level, the block is at a height $-h$ relative to the reference point (it's $h$ below the reference). The gravitational potential energy at any point is $U_g = mg(\text{height relative to reference})$, so at the table: $U_g = mg(-h) = -mgh$. Choice B ($+mgh$) is incorrect because it has the wrong sign—being below the reference point gives negative potential energy. When working with potential energy, always identify your reference point clearly and remember that positions below the reference have negative potential energy.

Question 4

1. $W_{\text{push}} = \Delta U_g + \Delta E_{\text{th}}$ (correct answer)
2. $W_{\text{push}} = \Delta K$ only, since speed is constant
3. $W_{\text{push}} + \Delta U_g = 0$ because mechanical energy is conserved
4. $W_{\text{push}} = \Delta U_g - \Delta E_{\text{th}}$

Explanation: This problem involves energy accounting when pushing a block up a rough incline at constant speed, requiring analysis of work done by all forces. At constant speed, kinetic energy doesn't change ($\Delta K = 0$), but the pusher must do work to both increase gravitational potential energy and compensate for energy lost to friction. The complete energy equation is: $W_{\text{push}} = \Delta K + \Delta U_g + \Delta E_{\text{th}} = 0 + \Delta U_g + \Delta E_{\text{th}}$, which simplifies to $W_{\text{push}} = \Delta U_g + \Delta E_{\text{th}}$. Choice D is incorrect because it subtracts thermal energy instead of adding it—the pusher must provide energy for both the increase in height and the energy dissipated by friction. When analyzing constant-speed motion with friction, remember that the applied force must do enough work to account for all energy increases, including thermal energy from friction.

Question 5

1. $mgh_1 = mgh_2 + \tfrac12 mv^2$ (correct answer)
2. $mgh_1 + \tfrac12 mv^2 = mgh_2$
3. $m\vec g\cdot \vec h_1 = m\vec g\cdot \vec h_2$
4. $mgh_1 = mgh_2$ because speed is not energy

Explanation: This problem requires energy conservation for a roller coaster with negligible friction, where the car has both potential and kinetic energy at different points. At height $h_1$, the car starts from rest with energy $E_1 = mgh_1 + 0$; at height $h_2$ with speed $v$, it has energy $E_2 = mgh_2 + \frac{1}{2}mv^2$. Since mechanical energy is conserved (no friction), $E_1 = E_2$, giving us: $mgh_1 = mgh_2 + \frac{1}{2}mv^2$. This shows that the initial potential energy equals the sum of final potential and kinetic energies. Choice D ($mgh_1 = mgh_2$) is incorrect because it ignores the kinetic energy at height $h_2$—the car has speed $v$, so it possesses kinetic energy that must be included. For energy conservation problems, always account for all forms of mechanical energy (both kinetic and potential) at each position.

Question 6

1. $v=\sqrt{gh}$
2. $v=\sqrt{2gh}$ (correct answer)
3. $v=2gh$
4. $v=\sqrt{2\vec g\cdot \vec h}$

Explanation: This problem requires conservation of energy for a pendulum bob falling through height $h$ with negligible air resistance. Since only gravity does work, mechanical energy is conserved between the initial position (height $h$, at rest) and the lowest point (height 0, speed $v$). The energy conservation equation is: $mgh + 0 = 0 + \frac{1}{2}mv^2$, where we choose the lowest point as our zero potential energy reference. Solving for $v$: $mgh = \frac{1}{2}mv^2$, cancel $m$ to get $gh = \frac{1}{2}v^2$, multiply by 2 to get $2gh = v^2$, then take the square root: $v = \sqrt{2gh}$. Choice A ($v = \sqrt{gh}$) is incorrect because it forgets the factor of 2 that comes from the $\frac{1}{2}$ in kinetic energy. Remember that when converting all potential energy to kinetic energy, the factor of $\frac{1}{2}$ in kinetic energy leads to a factor of 2 under the square root.

Question 7

1. $mgh = \tfrac12 mv^2$ because gravitational potential converts entirely to kinetic energy
2. $mgh - W_{\text{air}} = \tfrac12 mv^2$ (correct answer)
3. $mgh = \tfrac12 mv^2 - W_{\text{air}}$
4. $mgh = \tfrac12 m\vec v^2 + \vec W_{\text{air}}$

Explanation: This question explores conservation of energy with air resistance as a non-conservative force. The skier's initial gravitational potential energy is mgh, and air resistance does negative work -W_air during descent. The energy equation is initial potential plus work by air resistance equals final kinetic energy. Thus, mgh - W_air = ½ m v² accurately describes the situation. Choice C is incorrect as it subtracts W_air from the kinetic energy, which would imply air resistance increases potential energy. Remember to treat non-conservative work as reducing the mechanical energy available for conversion to kinetic energy in such scenarios.

Question 8

1. $v_2>v_1$ because friction adds thermal energy to increase speed
2. $v_2=v_1$ because both lose the same gravitational potential energy $mgh$
3. $v_2<v_1$ because friction reduces the kinetic energy at the bottom (correct answer)
4. $\vec v_2=\vec v_1-\vec W_f$

Explanation: This question evaluates the effect of friction on conservation of energy in ramp systems. For the frictionless ramp, all gravitational potential mgh converts to kinetic energy ½ m v₁². On the frictional ramp, friction dissipates energy as heat, reducing the final kinetic energy, so v₂ < v₁. This comparison holds because friction does negative work, leading to less speed. Choice A is incorrect as it claims friction adds thermal energy to increase speed, which contradicts energy dissipation. When comparing systems, calculate or reason about energy losses to predict qualitative outcomes like speed differences.

Question 9

1. $\tfrac12 kx^2 = \tfrac12 mv^2 + \mu_k mgL$ (correct answer)
2. $\tfrac12 kx^2 + \mu_k mgL = \tfrac12 mv^2$
3. $\tfrac12 kx^2 = \tfrac12 mv^2 - \mu_k mgL$
4. $\tfrac12 kx^2 = \tfrac12 m\vec v^2 + \mu_k mg\,\vec L$

Explanation: This question examines conservation of energy involving elastic potential, kinetic energy, and frictional work. The initial elastic potential energy is ½ k x², and friction does negative work -μ_k m g L over distance L. The energy balance is initial elastic potential plus work by friction equals final kinetic energy, leading to ½ k x² - μ_k m g L = ½ m v², or rearranged as ½ k x² = ½ m v² + μ_k m g L. This equation correctly captures the dissipation of energy due to friction. Choice C is incorrect because it subtracts the frictional term from the kinetic energy, reversing the energy loss. To solve similar problems, list all energy forms and subtract work by dissipative forces from the initial energy.

Question 10

1. $U_g=0$
2. $U_g=+mgH$
3. $U_g=-mgH$ (correct answer)
4. $U_g=-\dfrac{mg}{H}$

Explanation: This problem tests understanding of gravitational potential energy reference points. When we choose the zero of gravitational potential energy at height $y = H$ (the release point), the potential energy at any height $y$ is $U_g = mg(y - H)$. At the ground where $y = 0$, we have $U_g = mg(0 - H) = -mgH$. The negative sign indicates that the ground is below our chosen reference level. Choice B would be correct if we had chosen the ground as our zero reference point. When working with potential energy, always clearly identify your reference point and remember that potential energy can be negative when the object is below the reference level.

Question 11

1. $K = mgh - W_f$ (correct answer)
2. $K = mgh + W_f$
3. $K = m\vec g\cdot \vec h - W_f$
4. $K = mgh$ because total mechanical energy is conserved

Explanation: This problem requires energy accounting when both gravity and friction do work on a skier. The skier starts with gravitational potential energy $mgh$ (taking the bottom as zero reference) and zero kinetic energy, then friction does negative work $-W_f$ on the skier. The work-energy theorem states: $W_{\text{net}} = \Delta K$, where $W_{\text{net}} = W_{\text{gravity}} + W_{\text{friction}} = mgh + (-W_f) = mgh - W_f$. Since the skier starts from rest, $\Delta K = K_{\text{final}} - 0 = K_{\text{final}}$, so $K = mgh - W_f$. Choice B ($K = mgh + W_f$) is incorrect because it treats friction work as positive rather than negative—friction opposes motion and removes energy from the skier. When multiple forces do work, add their work algebraically, remembering that friction does negative work on moving objects.

Question 12

1. $\tfrac12 mv_0^2 = m\vec g\cdot \vec d$
2. $\Delta K + \Delta U_g + \Delta E_{\text{th}}=0$, with $\Delta E_{\text{th}}=\tfrac12 mv_0^2$ (correct answer)
3. $\tfrac12 mv_0^2 + U_g$ is conserved because the track is horizontal
4. $\Delta K = +f_k d$ because friction adds energy

Explanation: This problem involves energy conservation with friction, where a cart stops due to work done by friction. The correct energy accounting must include all energy changes: kinetic energy decreases to zero, gravitational potential energy doesn't change (horizontal track), and thermal energy increases due to friction. The complete energy conservation equation is $\Delta K + \Delta U_g + \Delta E_{\text{th}} = 0$, where $\Delta K = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$, $\Delta U_g = 0$ (horizontal), and $\Delta E_{\text{th}} = +\frac{1}{2}mv_0^2$ (positive because thermal energy increases). Choice D is incorrect because friction removes kinetic energy from the cart (does negative work on the cart), not adds it. When friction acts, always include thermal energy in your energy accounting—the lost mechanical energy becomes thermal energy.

Question 13

1. $h_{\max}=\dfrac{v_0^2}{g}$
2. $h_{\max}=\dfrac{v_0}{2g}$
3. $h_{\max}=\dfrac{v_0^2}{2g}$ (correct answer)
4. $h_{\max}=\dfrac{\vec v_0\cdot \vec v_0}{2\vec g}$

Explanation: This problem involves finding maximum height using energy conservation for projectile motion with negligible air resistance. At ground level, the ball has kinetic energy $\frac{1}{2}mv_0^2$ and zero potential energy; at maximum height, it has zero kinetic energy (momentarily at rest) and potential energy $mgh_{\max}$. The energy conservation equation is: $\frac{1}{2}mv_0^2 = mgh_{\max}$, where we use ground as our zero potential energy reference. Solving for $h_{\max}$: cancel $m$ to get $\frac{1}{2}v_0^2 = gh_{\max}$, then divide by $g$: $h_{\max} = \frac{v_0^2}{2g}$. Choice A ($h_{\max} = \frac{v_0^2}{g}$) is incorrect because it forgets the factor of $\frac{1}{2}$ from kinetic energy. For vertical motion problems, remember that maximum height occurs when all kinetic energy converts to potential energy, and the $\frac{1}{2}$ in kinetic energy appears in the final answer.

Question 14

1. $K+U_g$ is conserved because gravity is conservative
2. $K+U_g$ decreases because friction converts some energy to thermal energy (correct answer)
3. $K+U_g$ increases because friction does positive work
4. Mechanical energy is a vector, so its direction changes down the ramp

Explanation: This problem examines how mechanical energy changes when friction is present on a ramp. Mechanical energy is the sum of kinetic and gravitational potential energy: $E_{\text{mech}} = K + U_g$. When friction acts, it does negative work on the block, converting some mechanical energy into thermal energy through heat. The energy conservation equation including thermal energy is: $(K + U_g)_{\text{initial}} = (K + U_g)_{\text{final}} + E_{\text{thermal}}$, which shows that mechanical energy decreases by the amount converted to thermal energy. Choice A is incorrect because it ignores friction—mechanical energy is only conserved when no non-conservative forces (like friction) do work. When friction is present, mechanical energy always decreases as it gets converted to thermal energy, following the principle that total energy (including thermal) is conserved.

Question 15

1. $v=\sqrt{\dfrac{kx}{m}}$
2. $v=\sqrt{\dfrac{kx^2}{m}}$
3. $v=x\sqrt{\dfrac{k}{m}}$ (correct answer)
4. $v=\dfrac{kx^2}{2m}$

Explanation: This problem involves energy conservation with a spring launching a block on a frictionless surface. Initially, all energy is stored as elastic potential energy in the compressed spring ($U_s = \frac{1}{2}kx^2$), and finally all energy is kinetic when the spring reaches equilibrium. The energy conservation equation is: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$, where $x$ is the initial compression and $v$ is the final speed. Solving for $v$: cancel the $\frac{1}{2}$ to get $kx^2 = mv^2$, divide by $m$ to get $\frac{kx^2}{m} = v^2$, then take the square root: $v = x\sqrt{\frac{k}{m}}$. Choice B ($v = \sqrt{\frac{kx^2}{m}}$) is incorrect because it places $x^2$ inside the square root instead of factoring out $x$. When working with spring energy problems, remember that elastic potential energy is proportional to $x^2$, so one factor of $x$ comes outside the square root.

Question 16

1. $mgh - f_k d = \tfrac12 mv^2$ (correct answer)
2. $mgh + f_k d = \tfrac12 mv^2$
3. $mgh = \tfrac12 mv^2 + f_k\,\vec d$
4. $mg(h+d)=\tfrac12 mv^2$

Explanation: This question assesses the conservation of energy principle when non-conservative forces like friction are present. The initial gravitational potential energy of the block is mgh, which is partially converted to kinetic energy at the bottom and partially dissipated as thermal energy due to friction. The work done by friction is negative and equals -f_k d, so the energy equation is initial potential energy plus work by friction equals final kinetic energy. Thus, mgh - f_k d = ½ m v² correctly accounts for the energy transformation. Choice B is incorrect because it adds the frictional work instead of subtracting it, which would imply friction increases the kinetic energy. When applying conservation of energy with friction, always include the work done by non-conservative forces as part of the energy accounting to determine the final kinetic energy.

Question 17

1. $mgh = \tfrac12 kx^2$ (correct answer)
2. $mgh + \tfrac12 kx^2 = 0$
3. $mgh = \tfrac12 kx^2\,\hat{x}$
4. $mg(h-x)=\tfrac12 kx^2$

Explanation: This question tests conservation of energy in a system converting gravitational to elastic potential. The block starts with gravitational potential mgh and ends at maximum compression with elastic potential ½ k x² and zero kinetic energy. Since the track is frictionless, mechanical energy is conserved, so initial potential equals final elastic potential. Therefore, mgh = ½ k x² is the correct relation. Choice D is incorrect as it uses (h - x), assuming x affects height, which it does not in a horizontal spring. For energy transformations, identify points of interest and equate energies while accounting only for conservative forces if no dissipation occurs.

Question 18

1. $mgh = \tfrac12 mv^2 - f_k L$
2. $mgh = \tfrac12 mv^2 + f_k L$
3. $mgh - f_k L = \tfrac12 mv^2$ (correct answer)
4. $mgh = \tfrac12 m\vec v^2 + f_k L$

Explanation: This question evaluates understanding of conservation of energy with friction acting over a specific section of the path. The cart starts with gravitational potential energy mgh, which converts to kinetic energy, but friction does negative work -f_k L on the rough section. The conservation equation states that initial potential energy plus work by friction equals final kinetic energy after the rough section. Therefore, mgh - f_k L = ½ m v² properly relates the quantities. Choice B is incorrect as it adds f_k L to the kinetic energy, mistakenly treating frictional work as positive. A useful strategy is to identify conservative and non-conservative forces separately and ensure non-conservative work is subtracted when it opposes motion.

Question 19

1. $x=\dfrac{mgh}{k}$
2. $x=\sqrt{\dfrac{2mgh}{k}}$ (correct answer)
3. $x=\sqrt{\dfrac{mgh}{2k}}$
4. $x=\dfrac{2mgh}{k}$

Explanation: This problem requires applying conservation of energy to find the spring compression distance. Initially, the block has gravitational potential energy mgh at height h and zero kinetic energy (released from rest). At maximum compression, the block momentarily stops (zero kinetic energy) and all energy is stored as elastic potential energy ½kx² in the spring. Setting initial energy equal to final energy: mgh = ½kx². Solving for x gives x = √(2mgh/k). Choice A incorrectly omits the factor of ½ from the spring's potential energy formula. The key strategy is to identify energy forms at initial and final states, then apply conservation when no non-conservative forces do work.

Question 20

1. $mgh_1 + \tfrac12 mv_1^2 = mgh_2 + \tfrac12 mv_2^2$ (correct answer)
2. $mgh_1 - \tfrac12 mv_1^2 = mgh_2 - \tfrac12 mv_2^2$
3. $m g (h_1-h_2)=\tfrac12 m(\vec v_2-\vec v_1)^2$
4. $mgh_1 + \tfrac12 mv_1^2 = 0$ if $U=0$ at point 2

Explanation: This question tests conservation of mechanical energy in a frictionless roller coaster system. Total mechanical energy at point 1 is m g h₁ + ½ m v₁². At point 2, it is m g h₂ + ½ m v₂². Since energy is conserved with no friction or air resistance, these totals are equal: m g h₁ + ½ m v₁² = m g h₂ + ½ m v₂². Choice B is incorrect as it subtracts kinetic terms, which would not preserve energy equality. Always sum potential and kinetic energies at different points and set them equal for conservative systems to find relationships between variables.