Conservation of energy

Mechanical energy is the sum of the kinetic and potential energy, $E = K + U$, where $K = \tfrac{1}{2}mv^2$ and $U$ is the stored energy of position ($mgy$ near the ground, $\tfrac{1}{2}kx^2$ for a spring). Both are in joules (J).

The idea that runs through this whole topic is that $E = K + U$ stays constant as long as the only forces doing work are conservative, gravity and ideal springs. Then $K_i + U_i = K_f + U_f$: whatever leaves $U$ shows up in $K$ and vice versa.

Three habits cost points: assuming $E$ is conserved even when friction or drag is doing work (§2), misreading where the motion turns around on an energy graph (§3), and treating energy as something that gets used up instead of traded between $K$ and $U$ (§4).

Throughout this lesson, $g = 10$ m/s$^2$ where it is needed.

Conservation of $E = K + U$ has a condition attached: it holds only when the non-conservative forces do no work. Friction and air drag are non-conservative; they oppose the motion and turn mechanical energy into heat, so when they act the total $E$ does not stay fixed.

The bookkeeping is one line: the change in mechanical energy equals the work done by the non-conservative forces, $\Delta E = W_{nc}$, or $E_f = E_i + W_{nc}$. For friction and drag $W_{nc}$ is negative, so $E_f < E_i$: the system ends with less mechanical energy than it started with.

For example, a cart starts with $E_i = 16$ J and friction does $-6$ J over the run. Then $E_f = E_i + W_{nc} = 16 + (-6) = 10$ J. The $6$ J did not vanish; it became heat, which is no longer part of $K + U$.

The trap is to use $K_i + U_i = K_f + U_f$ whenever you see an energy problem. Check first: is anything but gravity or an ideal spring doing work? If friction or drag is, write $E_f = E_i + W_{nc}$ instead.

Draw $U(x)$ as a curve and the total energy as a horizontal line at height $E$. Because $E = K + U$ and $K$ can never be negative, the particle can only be where the curve sits at or below the line: the allowed region is $U \le E$. Where $U > E$ is forbidden.

A turning point is where the curve meets the line, $U = E$. There the kinetic energy is $K = E - U = 0$, so the particle momentarily stops and reverses. It is not where $K$ is largest; it is where $K = 0$.

For example, take $U(x) = 2x^2$ with the energy line at $E = 18$ J. The turning points are where $2x^2 = 18$, so $x = \pm 3$ m. At those points $K = 0$; between them $K = E - U > 0$ and the particle moves.

The trap is to read the turning point as the place with the most energy and call $K = E$ there. It is the opposite: at the turning point all the energy is potential, $U = E$, and $K = 0$.

The kinetic energy at any point is the gap between the energy line and the curve, $K = E - U$. Where $U$ is low the gap is wide and the particle moves fast; where $U$ rises toward the line the gap shrinks and it slows. Speed is greatest where $U$ is lowest.

Nothing is being used up. With no friction the total $E$ is fixed, and the particle simply trades energy back and forth: climbing converts $K$ into $U$, descending converts $U$ back into $K$. A pendulum or a cart in a valley does this on every pass.

So "the energy runs out at the top, then comes back" is the wrong picture. At the top of a swing $K$ is zero but $U$ is at its largest; the total is the same as at the bottom. The split changes; the sum does not.

Reading $K$ off the graph is just subtraction: find $U$ at the point, subtract it from $E$, and what is left is $K = E - U$. If the line sits at $E = 18$ J and the curve reads $U = 6$ J there, then $K = 12$ J.

A conservation-of-energy question comes down to one judgment call and two graph reads: whether $E$ is conserved, where the motion turns around, and how fast it moves in between.

Does conservation hold? Ask what does work. If only gravity and ideal springs do, $E = K + U$ is constant and $K_i + U_i = K_f + U_f$. If friction or drag does work, switch to $E_f = E_i + W_{nc}$ with $W_{nc} < 0$.

The turning point. On a $U(x)$ graph the motion lives where $U \le E$ and turns around where the curve meets the line, $U = E$, because $K = E - U = 0$ there. The turning point is where $K$ is zero, not where it is largest.

The speed. Kinetic energy is the gap $K = E - U$: fastest where $U$ is lowest, slowest where $U$ is highest. Energy is traded between $K$ and $U$, not used up, so the total stays fixed when no friction acts.

Setup. A $2$ kg cart is released from rest on a track ($g = 10$ m/s$^2$), taking the bottom of the track as the zero of potential energy. (a) On a frictionless track it starts where $U = 16$ J; find its speed at the bottom. (b) On a $U(x)$ graph the total energy is a line at $E = 16$ J; describe where the cart turns around and where it is fastest. (c) Now friction does $-6$ J over the run; find the speed at the bottom.

(a) Conservation. Only gravity does work, so $E = K + U$ is constant at $16$ J. At the start $K_i = 0$, so $E = 16$ J. At the bottom $U = 0$, so $K = E - U = 16$ J. Then $\tfrac{1}{2}(2)v^2 = 16$ gives $v^2 = 16$ and $v = 4$ m/s.

(b) The graph. The cart can be only where $U \le 16$ J. It turns around where the curve meets the line, $U = 16$ J, and there $K = E - U = 0$. It is fastest at the lowest point of the curve, where the gap $K = E - U$ is widest, not on the energy line itself.

(c) With friction. Friction is non-conservative, so $E$ is not conserved: $E_f = E_i + W_{nc} = 16 + (-6) = 10$ J. At the bottom $U = 0$, so $K = 10$ J, and $\tfrac{1}{2}(2)v^2 = 10$ gives $v = \sqrt{10} \approx 3.2$ m/s, slower than the frictionless $4$ m/s, and the cart cannot climb as high on the far side.

The traps to avoid. Do not assume $K_i + U_i = K_f + U_f$ when friction acts (use $E_f = E_i + W_{nc}$), do not read the turning point as $K = E$ (it is $K = 0$, with $U = E$), and do not picture energy as used up (it is fastest where $U$ is lowest because $K = E - U$ is largest there).