Potential Energy - AP Physics C Mechanics

1. In the described system, how much does the potential energy change as a $4.00\,\text{kg}$ block descends $h=1.50\,\text{m}$ on a frictionless incline ($g=9.80\,\text{m/s}^2$)?
   Answer: -58.8J This question tests understanding of potential energy changes in gravitational systems, specifically when an object descends and loses height. When an object moves downward in a gravitational field, its potential energy decreases, resulting in a negative change: ΔU = mgΔh, where Δh is negative for downward motion. In this scenario, a 4.00 kg block descends 1.50 m, so Δh = -1.50 m, giving ΔU = (4.00)(9.80)(-1.50) = -58.8 J. Choice A is correct because it properly accounts for the negative change in height and the resulting negative change in potential energy. Choice B (58.8 J) is incorrect because it ignores the direction of motion, treating the descent as if it were an ascent, a common sign error students make. To help students: Emphasize that potential energy changes depend on the direction of motion - decreasing height means decreasing potential energy. Use energy conservation examples to show that lost potential energy becomes kinetic energy, reinforcing the negative sign's physical meaning.
2. In the described system, how much does the potential energy change as a $1.20\,\text{kg}$ block descends $h=2.50\,\text{m}$ on a frictionless incline ($g=9.80\,\text{m/s}^2$)?
   Answer: -29.4J This question tests understanding of potential energy changes during downward motion on an incline. When an object descends, its gravitational potential energy decreases regardless of the path taken - only the vertical height change matters. In this scenario, a 1.20 kg block descends a vertical height of 2.50 m, so ΔU = mgΔh = (1.20)(9.80)(-2.50) = -29.4 J. Choice A is correct because it properly includes the negative sign indicating a decrease in potential energy. Choice B (29.4 J) is incorrect because it ignores the direction of motion, treating the energy change as positive when the object is actually losing potential energy. To help students: Emphasize that the incline angle is irrelevant for potential energy calculations - only vertical displacement matters. Use energy conservation to show that this lost potential energy becomes kinetic energy, making the negative sign physically meaningful.
3. A particle's motion is constrained to the x-axis and is influenced by a potential energy function $U(x)$. If the particle moves from an initial position $x_i$ to a final position $x_f$, the work done by the conservative force associated with $U(x)$ is equal to which of the following?
   Answer: $U(x_i) - U(x_f)$ By definition, the work done by a conservative force $W_c$ is equal to the negative of the change in potential energy, $\Delta U$. The change in potential energy is $\Delta U = U_{final} - U_{initial} = U(x_f) - U(x_i)$. Therefore, $W_c = -\Delta U = -(U(x_f) - U(x_i)) = U(x_i) - U(x_f)$.
4. The potential energy of a particle moving in the xy-plane is given by $U(x, y) = 3x^2y - 5y^3$. What is the force vector $\vec{F}$ acting on the particle at the point (2, 1)?
   Answer: $-12\hat{i} + 3\hat{j}$ The force vector is the negative gradient of the potential energy: $\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})$. First, find the partial derivatives: $\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(3x^2y - 5y^3) = 6xy$. $\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(3x^2y - 5y^3) = 3x^2 - 15y^2$. Now evaluate at the point (2, 1): $\frac{\partial U}{\partial x}|_{(2,1)} = 6(2)(1) = 12$. $\frac{\partial U}{\partial y}|_{(2,1)} = 3(2^2) - 15(1^2) = 12 - 15 = -3$. So, $\vec{F} = -(12\hat{i} - 3\hat{j}) = -12\hat{i} + 3\hat{j}$
5. A particle is subject to a conservative force described by $F(x) = -ax^3$, where $a$ is a positive constant. If the potential energy is taken to be zero at $x=0$, what is the potential energy function $U(x)$?
   Answer: $\frac{1}{4}ax^4$ The potential energy function $U(x)$ is found by integrating the negative of the force: $U(x) - U(x_0) = -\int_{x_0}^x F(x') dx'$. With $U(0)=0$, we have $U(x) = -\int_0^x (-ax'^3) dx' = a \int_0^x x'^3 dx' = a [\frac{1}{4}x'^4]_0^x = \frac{1}{4}ax^4$.
6. A uniform disk of radius $R$ and mass $M$ rolls without slipping down an inclined plane of height $h$. When the disk reaches the bottom, what fraction of its total kinetic energy is in the form of rotational kinetic energy?
   Answer: $\frac{1}{3}$ When you encounter rolling motion problems on AP Physics C, you're dealing with objects that have both translational and rotational kinetic energy. The key insight is understanding how these energies relate through the no-slip condition. For a rolling disk, the no-slip condition gives us $v = \omega R$, where $v$ is the center-of-mass velocity and $\omega$ is the angular velocity. The total kinetic energy consists of translational kinetic energy $KE_{trans} = \frac{1}{2}Mv^2$ and rotational kinetic energy $KE_{rot} = \frac{1}{2}I\omega^2$. For a uniform disk, the moment of inertia is $I = \frac{1}{2}MR^2$. Substituting the no-slip condition $\omega = \frac{v}{R}$: $KE_{rot} = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}Mv^2$ The total kinetic energy is: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2$ Therefore, the fraction of rotational energy is: $\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{4}Mv^2}{\frac{3}{4}Mv^2} = \frac{1}{3}$ Choice A ($\frac{1}{4}$) represents the ratio of rotational to translational energy, not to total energy. Choice B ($\frac{1}{2}$) would apply to a solid sphere, which has $I = \frac{2}{5}MR^2$. Choice D ($\frac{2}{3}$) gives the translational fraction instead of rotational. Remember: For rolling motion problems, always identify the moment of inertia first, then use the no-slip condition to relate linear and angular quantities. The energy fractions depend entirely on the object's shape through its moment of inertia.
7. A particle moves along the x-axis under the influence of a conservative force. Its potential energy is described by the function $U(x) = \frac{A}{x^2} - \frac{B}{x}$, where A and B are positive constants. What is the force $F(x)$ exerted on the particle at position $x$?
   Answer: $F(x) = \frac{2A}{x^3} - \frac{B}{x^2}$ The force is the negative gradient of the potential energy. For one dimension, this is $F(x) = -\frac{dU}{dx}$. Differentiating $U(x)$ with respect to $x$ gives $\frac{dU}{dx} = \frac{d}{dx}(Ax^{-2} - Bx^{-1}) = -2Ax^{-3} - (-1)Bx^{-2} = -\frac{2A}{x^3} + \frac{B}{x^2}$. Therefore, $F(x) = -(-\frac{2A}{x^3} + \frac{B}{x^2}) = \frac{2A}{x^3} - \frac{B}{x^2}$.
8. Two point masses $m_1$ and $m_2$ are initially separated by distance $r_i$. They are then moved to a final separation of $r_f$, where $r_f > r_i$. During this process, the change in gravitational potential energy is $\Delta U$. If the same masses are moved from separation $2r_i$ to separation $2r_f$, what is the change in gravitational potential energy for this second process?
   Answer: $\frac{\Delta U}{2}$ The gravitational potential energy between two point masses is $U = -\frac{Gm_1m_2}{r}$. For the first process: $\Delta U = -\frac{Gm_1m_2}{r_f} - \left(-\frac{Gm_1m_2}{r_i}\right) = Gm_1m_2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$. For the second process with doubled distances: $\Delta U_2 = -\frac{Gm_1m_2}{2r_f} - \left(-\frac{Gm_1m_2}{2r_i}\right) = Gm_1m_2\left(\frac{1}{2r_i} - \frac{1}{2r_f}\right) = \frac{1}{2}Gm_1m_2\left(\frac{1}{r_i} - \frac{1}{r_f}\right) = \frac{\Delta U}{2}$. Choice A would be correct if gravitational PE were proportional to distance rather than inverse distance. Choice C incorrectly doubles the change. Choice D would result from incorrectly applying an inverse square relationship.
9. Two identical masses are connected by a massless rope that passes over a massless pulley. One mass hangs vertically, while the other slides on a frictionless inclined plane that makes an angle $\theta$ with the horizontal. The system is released from rest when both masses are at the same height. After the hanging mass has descended a distance $h$, what is the change in total potential energy of the system?
   Answer: $-mgh(1 - \sin\theta)$ The hanging mass loses potential energy $mgh$ as it descends distance $h$. The mass on the incline gains potential energy as it moves up the incline. Since the rope is inextensible, when the hanging mass descends $h$, the mass on the incline moves distance $h$ along the incline, gaining height $h\sin\theta$. Thus it gains potential energy $mgh\sin\theta$. The total change in potential energy is $-mgh + mgh\sin\theta = -mgh(1 - \sin\theta)$. Choice A incorrectly adds the contributions instead of subtracting. Choice C ignores the change in potential energy of the hanging mass. Choice D uses an incorrect geometric relationship.
10. A particle of mass $m$ moves in a region where the potential energy varies as $U(x) = U_0\cos\left(\frac{\pi x}{L}\right)$, where $U_0 > 0$ and $L > 0$. The particle has total mechanical energy $E = \frac{U_0}{2}$. At which positions can the particle be found?
    Answer: Only at positions where $\cos\left(\frac{\pi x}{L}\right) \leq \frac{1}{2}$ When analyzing particle motion in a potential energy field, you need to apply conservation of mechanical energy: the particle can only exist at positions where its kinetic energy remains non-negative ($KE = E - U(x) \geq 0$). Given that the total mechanical energy is $E = \frac{U_0}{2}$ and the potential energy is $U(x) = U_0\cos\left(\frac{\pi x}{L}\right)$, the kinetic energy at any position is: $KE = E - U(x) = \frac{U_0}{2} - U_0\cos\left(\frac{\pi x}{L}\right)$ For the particle to physically exist at a position, we need $KE \geq 0$: $\frac{U_0}{2} - U_0\cos\left(\frac{\pi x}{L}\right) \geq 0$ Dividing by $U_0$ and rearranging: $\cos\left(\frac{\pi x}{L}\right) \leq \frac{1}{2}$ This confirms answer D is correct. Let's examine why the other options fail: A suggests the particle exists only where $\cos\left(\frac{\pi x}{L}\right) = \frac{1}{2}$, which would mean zero kinetic energy everywhere – this describes only the turning points, not the entire allowed region. B requires $\cos\left(\frac{\pi x}{L}\right) \geq \frac{1}{2}$, which would make kinetic energy negative (impossible). C claims the particle can exist everywhere from $0 \leq x \leq L$, ignoring the energy constraint entirely. Study tip: In potential energy problems, always check that $E - U(x) \geq 0$ to find classically allowed regions. The particle is "trapped" where its total energy exceeds the potential energy, creating natural boundaries called turning points.
11. A non-ideal spring exerts a restoring force given by $F(x) = -kx - \beta x^3$, where $k$ and $\beta$ are positive constants and $x$ is the displacement from equilibrium. What is the potential energy $U(x)$ stored in the spring when it is displaced a distance $x$, assuming $U(0) = 0$?
    Answer: $\frac{1}{2}kx^2 + \frac{1}{4}\beta x^4$ The potential energy $U(x)$ is found by integrating the negative of the force: $U(x) - U(0) = -\int_0^x F(x') dx'$. With $U(0)=0$, we have $U(x) = -\int_0^x (-kx' - \beta x'^3) dx' = \int_0^x (kx' + \beta x'^3) dx' = [\frac{1}{2}kx'^2 + \frac{1}{4}\beta x'^4]_0^x = \frac{1}{2}kx^2 + \frac{1}{4}\beta x^4$.
12. A particle moves in a one-dimensional potential given by $U(x) = x^4 - 8x^2 + 12$. If the particle has a total mechanical energy of $E = 4$ J, which of the following is a turning point of its motion?
    Answer: $x = 2$ m Turning points occur where the kinetic energy is zero, so the total energy equals the potential energy: $E = U(x)$. We set $4 = x^4 - 8x^2 + 12$. This gives the equation $x^4 - 8x^2 + 8 = 0$. Let $y = x^2$, then $y^2 - 8y + 8 = 0$. Using the quadratic formula: $y = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2}$. Since $y = x^2 \geq 0$, both solutions are valid: $x^2 = 4 + 2\sqrt{2} \approx 6.83$ or $x^2 = 4 - 2\sqrt{2} = 4$. The turning points are at $x = \pm 2$ m and $x = \pm\sqrt{4 + 2\sqrt{2}}$ m. Of the choices given, $x = 2$ m is a turning point.
13. A particle of mass $m$ is located inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center ($r < R$). The gravitational force on the particle is $F_g(r) = -G\frac{Mmr}{R^3}$. If the potential energy is defined to be zero at the center of the sphere ($U(0)=0$), what is the potential energy $U(r)$ of the particle-sphere system?
    Answer: $\frac{GMmr^2}{2R^3}$ The change in potential energy is the negative of the work done by the conservative force. $U(r) - U(0) = -\int_0^r F_g(r') dr' = -\int_0^r (-G\frac{Mmr'}{R^3}) dr' = G\frac{Mm}{R^3} \int_0^r r' dr'$. Evaluating the integral gives $G\frac{Mm}{R^3} [\frac{1}{2}r'^2]_0^r = G\frac{Mm r^2}{2R^3}$. Since $U(0)=0$, we have $U(r) = \frac{GMmr^2}{2R^3}$.
14. The potential energy of a particle moving along the x-axis is given by the function $U(x) = x^3 - 6x^2 + 9x + 5$. At which position does the particle experience stable equilibrium?
    Answer: $x = 3$ m Equilibrium occurs where the net force is zero, which means $F(x) = -\frac{dU}{dx} = 0$. Here, $\frac{dU}{dx} = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. So, equilibrium points are at $x=1$ and $x=3$. For stable equilibrium, the potential energy must be at a local minimum, which requires $\frac{d^2U}{dx^2} > 0$. The second derivative is $\frac{d^2U}{dx^2} = 6x - 12$. At $x=1$, $\frac{d^2U}{dx^2} = -6$ (unstable). At $x=3$, $\frac{d^2U}{dx^2} = 6$ (stable). Thus, stable equilibrium is at $x=3$ m.
15. The gravitational potential energy of an object of mass $m$ at a height $h$ above the ground is $U = mgh$. This definition assumes the potential energy is zero at ground level ($h=0$). If the zero point for potential energy is redefined to be at a height $h_0$ above the ground, how does this affect the calculated work done by gravity as the object moves from height $h_1$ to $h_2$?
    Answer: The work done by gravity is unchanged. The work done by a conservative force, like gravity, equals the negative change in potential energy: $W_g = -\Delta U$. With the original definition, $W_g = -(mgh_2 - mgh_1) = mg(h_1 - h_2)$. With the new definition, $U' = mg(h - h_0)$, so the change is $\Delta U' = mg(h_2 - h_0) - mg(h_1 - h_0) = mgh_2 - mgh_1$. The work done is $W'_g = -\Delta U' = -(mgh_2 - mgh_1) = mg(h_1 - h_2)$. The work done is independent of the choice of the zero point for potential energy.
16. A block of mass $m$ is attached to a spring with spring constant $k$. The spring is stretched a distance $x$, storing potential energy $U$. If the spring is cut into two equal halves and the block is attached to one of the halves, what distance must the half-spring be stretched to store the same potential energy $U$?
    Answer: $x/\sqrt{2}$ When a spring of constant $k$ is cut in half, each half has a new spring constant $k' = 2k$. The original potential energy is $U = \frac{1}{2}kx^2$. We want the new potential energy $U'$ to be equal to $U$. Let the new stretch distance be $x'$. Then $U' = \frac{1}{2}k'(x')^2 = \frac{1}{2}(2k)(x')^2 = k(x')^2$. Setting $U' = U$ gives $k(x')^2 = \frac{1}{2}kx^2$. Solving for $x'$ yields $(x')^2 = \frac{x^2}{2}$, so $x' = \frac{x}{\sqrt{2}}$.
17. A particle is moved from point P to point Q in a region where a conservative force field exists. Path 1 is a direct straight line from P to Q. Path 2 is a semicircular arc from P to Q. The work done by the conservative force is $W_1$ along Path 1 and $W_2$ along Path 2. Which of the following statements is always true?
    Answer: $W_1 = W_2$ A defining property of a conservative force is that the work it does on a particle moving between two points is independent of the path taken. Since the force is conservative and both paths start at P and end at Q, the work done must be the same for both paths. Therefore, $W_1 = W_2$.
18. A uniform rod of length $L$ and mass $M$ is pivoted at one end and can rotate freely in a vertical plane. The rod is initially horizontal and released from rest. When the rod has rotated through an angle $\theta$ from the horizontal, what is the gravitational potential energy of the rod relative to its initial position?
    Answer: $-\frac{MgL}{2}\sin\theta$ The center of mass of a uniform rod is at its geometric center, which is at distance $L/2$ from the pivot point. Initially, when the rod is horizontal, the center of mass is at height $L/2$ above the lowest possible position of the center of mass. When the rod rotates through angle $\theta$ from horizontal (measuring downward), the center of mass descends by a vertical distance $\frac{L}{2}\sin\theta$. Therefore, the change in gravitational potential energy is $\Delta U = -Mg \cdot \frac{L}{2}\sin\theta = -\frac{MgL}{2}\sin\theta$. Choice B would be correct if $\theta$ were measured from the vertical position. Choice C has the wrong trigonometric function. Choice D is missing the factor of 1/2 that accounts for the center of mass being at the middle of the rod.
19. A potential energy function can be defined for a force $F$ if and only if the work done by the force is independent of the path taken between any two points. For which of the following forces can a potential energy function NOT be defined?
    Answer: The drag force on an object moving through a fluid. A potential energy function can only be defined for conservative forces, for which the work done is path-independent. Gravitational, electrostatic, and ideal spring forces are all conservative. The drag force, however, is a non-conservative force. It always opposes the direction of motion, so the work it does depends on the total length of the path taken, not just the start and end points. Therefore, a potential energy function cannot be defined for the drag force.
20. A block slides down a frictionless curved track and then moves horizontally before encountering a spring. The block compresses the spring by a maximum distance $d$ before coming momentarily to rest. If the initial height of the block above the horizontal surface is $h$, and the spring constant is $k$, what is the relationship between the elastic potential energy stored in the spring and the initial gravitational potential energy of the block?
    Answer: The elastic potential energy equals the initial gravitational potential energy By conservation of mechanical energy, the initial gravitational potential energy $mgh$ is completely converted to elastic potential energy $\frac{1}{2}kd^2$ when the spring is maximally compressed (since the block is momentarily at rest). Therefore, $mgh = \frac{1}{2}kd^2$, meaning the elastic potential energy equals the initial gravitational potential energy. Choice B incorrectly suggests there's additional kinetic energy, but at maximum compression the block is at rest. Choice C is wrong because the track is frictionless, so no energy is lost. Choice D is incorrect because energy conservation provides a definite relationship regardless of the block's mass.

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