Conservation of Energy - AP Physics 1

A block of mass $m$ is released from rest at height $h$ above a table. A student sets $U_g=0$ at the release point. What is $U_g$ at the table level?
Answer: $-mgh$ 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.
Two identical balls are dropped from rest: Ball 1 from height $h$ and Ball 2 from height $2h$ . Air resistance is negligible. Just before hitting the ground, each ball has speed $v_1$ and $v_2$ , respectively. What is the ratio $v_2/v_1$ ?
Answer: $\sqrt{2}$ This problem applies conservation of energy to free fall from different heights. For ball 1 dropped from height h: mgh = ½mv₁², giving v₁ = √(2gh). For ball 2 dropped from height 2h: mg(2h) = ½mv₂², giving v₂ = √(4gh) = 2√(gh). The ratio is v₂/v₁ = 2√(gh)/√(2gh) = 2/√2 = √2. Choice A incorrectly assumes speed is proportional to height rather than to the square root of height. The key insight is that kinetic energy is proportional to height, but speed is proportional to the square root of height.
Two identical blocks start from rest at the same height $h$ on two different ramps. Ramp 1 is frictionless; Ramp 2 has kinetic friction that does negative work of magnitude $W_f$ on the block. Air resistance is negligible. Both reach the bottom. Which comparison of their bottom speeds is correct?
Answer: $v_2<v_1$ because friction reduces the kinetic energy at the bottom 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.
A block slides down a rough ramp from height $h$ ; friction is present and the block reaches the bottom with speed $v$ . Which energy accounting is correct?
Answer: $mgh=\tfrac12 mv^2+E_{\text{th}}$ with $E_{\text{th}}>0$ This problem involves energy conservation with friction creating thermal energy. The initial gravitational potential energy $mgh$ is converted into two forms: kinetic energy $\frac{1}{2}mv^2$ of the block and thermal energy $E_{th}$ due to friction. The correct energy equation is: $mgh = \frac{1}{2}mv^2 + E_{th}$ where $E_{th} > 0$ represents the positive thermal energy generated. Choice A incorrectly assumes no energy is lost to friction, which would only be true for a frictionless surface. When friction is present in energy problems, always account for the thermal energy produced, which equals the mechanical energy lost from the system.
A ball of mass $m$ is thrown straight upward from height $y_0$ above the ground with initial speed $v_0$ . Air resistance is negligible. A student chooses gravitational potential energy to be zero at the launch point instead of at the ground. Which statement about the ball's maximum height above the ground is correct?
Answer: It is independent of the chosen zero level for potential energy This problem tests understanding of gravitational potential energy reference levels in conservation of energy. The ball's maximum height above ground depends only on its initial kinetic energy ½mv₀² converting to gravitational potential energy. Using energy conservation: ½mv₀² = mg(h_max - y₀), giving h_max = y₀ + v₀²/(2g). This result is independent of where we choose $U_g$ = 0 because only differences in potential energy matter in conservation equations. Choice A incorrectly suggests the physical result depends on our arbitrary choice of reference level. The key insight is that energy differences, not absolute values, determine physical outcomes.
A block of mass $m$ slides down a vertical drop of height $h$ and then across a rough horizontal surface, coming to rest after distance $d$ . The only nonconservative force is kinetic friction on the horizontal surface; air resistance is negligible. Which expression gives the magnitude of the friction force $f_k$ ?
Answer: $f_k=\dfrac{mgh}{d}$ This problem requires finding the friction force using conservation of energy. The block starts with gravitational potential energy mgh at the top and zero kinetic energy. On the horizontal surface, friction does negative work $W_f$ = - $f_k$ ·d, bringing the block to rest. The energy equation is: mgh + (- $f_k$ ·d) = 0, which gives $f_k$ = mgh/d. Choice A incorrectly uses the kinetic energy at the bottom instead of recognizing that all initial potential energy is dissipated by friction. The key insight is that the work done by friction equals the total mechanical energy lost, which equals the initial potential energy.
A roller coaster car of mass $m$ moves along a track with negligible friction. Point $1$ is at height $h_1$ with speed $v_1$ , and point $2$ is at height $h_2$ with speed $v_2$ . Air resistance is negligible. Which relation between these quantities must be true?
Answer: $mgh_1 + \tfrac12 mv_1^2 = mgh_2 + \tfrac12 mv_2^2$ 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.
A block slides down a ramp with kinetic friction and reaches the bottom with speed $v$ . Which statement about mechanical energy is correct?
Answer: $K+U_g$ decreases because friction converts some energy to thermal energy 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.
A roller coaster car (mass $m$ ) starts from rest at height $h_1$ and reaches height $h_2$ with speed $v$ ; friction is negligible. Which relation is correct?
Answer: $mgh_1 = mgh_2 + \tfrac12 mv^2$ 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.
A cart of mass $m$ starts from rest at height $h$ above a reference level and rolls down a track. The track is smooth (friction negligible) until the cart enters a rough horizontal section of length $L$ , where kinetic friction of magnitude $f_k$ acts. The cart exits the rough section with speed $v$ . Which equation best represents conservation of energy with work by friction included?
Answer: $mgh - f_k L = \tfrac12 mv^2$ 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.
A block slides down a rough incline, dropping vertical distance $h$ . It starts from rest and reaches speed $v$ . What is the thermal energy increase $\Delta E_{\text{th}}$ due to friction?
Answer: $\Delta E_{\text{th}}=mgh-\tfrac12 mv^2$ This question assesses calculating thermal energy increase from friction on a rough incline using energy conservation principles. The initial potential energy is mgh, and the final kinetic energy is (1/2) m v². The difference, ΔE_th = mgh - (1/2) m v², represents the energy converted to thermal form due to friction. This arises because friction does negative work, reducing mechanical energy. Option A is wrong as it reverses the terms, yielding a negative value which doesn't make sense for energy increase. For dissipative systems, subtract final mechanical energy from initial to find energy lost to thermal or other forms.
A pendulum bob drops from height $h$ above its lowest point; air resistance is negligible. What is $v$ at the bottom?
Answer: $v=\sqrt{2gh}$ 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.
A roller-coaster car of mass $m$ moves on a track with negligible friction. At point A it is at height $h_A$ with speed $v_A$ , and at point B it is at height $h_B$ with speed $v_B$ . Take $U_g=0$ at the ground. Which relation must be true?
Answer: $mgh_A+\tfrac12 mv_A^2=mgh_B+\tfrac12 mv_B^2$ This problem tests conservation of mechanical energy for a frictionless roller coaster. With no non-conservative forces (friction is negligible), total mechanical energy remains constant between any two points. At point A, the total energy is gravitational potential energy mgh_A plus kinetic energy ½mv_A². At point B, it's mgh_B + ½mv_B². Conservation of energy requires these totals to be equal: mgh_A + ½mv_A² = mgh_B + ½mv_B². Choice D incorrectly uses momentum (mv) instead of kinetic energy (½mv²) in the conservation equation. The strategy is to recognize when mechanical energy is conserved (no friction or air resistance) and write the conservation equation.
A ball is thrown upward from ground with speed $v_0$ ; air resistance is negligible. What is the maximum height above the ground?
Answer: $h_{\max}=\dfrac{v_0^2}{2g}$ 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.
A block of mass $m$ is released from rest at the top of a frictionless track at height $h$ above the bottom. At the bottom, it compresses a horizontal spring of constant $k$ by a maximum amount $x$ . Air resistance is negligible. Which equation correctly describes the energy transformation at maximum compression?
Answer: $mgh = \tfrac12 kx^2$ 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.
A skier of mass $m$ descends a vertical drop $h$ and experiences air drag that dissipates energy $E_d$ . Starting from rest, what is the skier's speed at the bottom?
Answer: $v=\sqrt{\dfrac{2(mgh-E_d)}{m}}$ 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.
A block is launched up a frictionless ramp from point $A$ with speed $v_A$ and rises to point $B$ . If $y_A$ and $y_B$ are vertical positions, what relation is correct?
Answer: $\tfrac12 mv_A^2+mg y_A=\tfrac12 mv_B^2+mg y_B$ This question tests conservation of mechanical energy on a frictionless ramp. The total energy at A is (1/2) m $v_A$ ² + m g $y_A$ . At B, it is (1/2) m $v_B$ ² + m g $y_B$ , and since energy is conserved, they are equal. This holds as no non-conservative work is done. Option B is incorrect with negative potential terms, which might stem from mishandling signs in energy equations. Always set total mechanical energy equal at different points in conservative systems to relate speeds and heights.
A block of mass $m$ is released from rest at height $h$ above the floor and slides down a rough ramp to the floor. The ramp exerts a constant kinetic friction force of magnitude $f_k$ over a distance $d$ along the ramp. Air resistance is negligible. The block reaches the bottom with speed $v$ . Which energy-accounting equation correctly relates these quantities?
Answer: $mgh - f_k d = \tfrac12 mv^2$ 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.
A block ( $m$ ) is pushed up a rough incline a distance $L$ at constant speed; friction is present. Which energy accounting is correct for the block?
Answer: $W_{\text{push}} = \Delta U_g + \Delta E_{\text{th}}$ 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.
A block-spring system oscillates on a horizontal frictionless surface. When the spring is stretched by $A$ , the block is momentarily at rest. What is the total mechanical energy $E$ of the system?
Answer: $E=\tfrac12 kA^2$ This question examines total mechanical energy in a horizontal spring-block oscillator. At maximum stretch A, kinetic energy is zero, so total E = (1/2) k A², the elastic potential energy. This energy remains constant throughout oscillation on the frictionless surface. No gravitational term is needed due to the horizontal setup. Option D is wrong as it adds an irrelevant m g A term, perhaps confusing with vertical systems. For oscillatory motion, calculate total energy at an extreme position where kinetic energy is zero for simplicity.

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