Mistake Master
Mistake Master
Kinetic energy belongs to one moving object. Potential energy belongs to a system whose arrangement stores work waiting to happen. A ball above the ground; a compressed spring. Two formulas cover the course, and three pitfalls catch almost everyone: forgetting it is a system property, forgetting it scales as the square of the spring's compression, and forgetting the zero of $U$ is a free choice.
A boulder at the top of a hill is not moving, but a lot is going to happen if you let it go. A spring squeezed between two blocks is sitting still, but it will launch the blocks the instant the bracket is removed. In both cases the system holds potential energy: energy stored in the way the parts are arranged, ready to convert into kinetic energy when the arrangement is allowed to change.
The rest is two formulas, three traps to dodge, and a ten-scenario check at the bottom.
Gravitational potential energy, near Earth. When a mass $m$ moves between two heights near Earth's surface, the gravitational potential energy of the ball-Earth system changes by
$$\Delta U_g = m g\, \Delta y.$$Two things to notice. The formula gives the change $\Delta U_g$ between two heights. The absolute value of $U_g$ at any one height depends on where you set $U_g = 0$, and that is your call. Common choices are the floor, the table surface, or wherever the object starts. None is more correct than another.
Gravitational potential energy, two masses far apart. When two masses interact at long range (planet and satellite, Earth and Moon), the system potential energy is
$$U_G = -\dfrac{G m_1 m_2}{r}.$$Negative because the convention sets $U_G = 0$ at infinite distance; anything bound by gravity has $U_G < 0$. AP Physics 1 rarely plugs numbers into this. The negative sign is real.
Elastic potential energy. An ideal Hooke's-law spring displaced by $\Delta x$ from its relaxed length stores
$$U_s = \tfrac{1}{2} k (\Delta x)^2.$$Always non-negative. Compression and stretch by the same amount store the same energy because $(\Delta x)^2$ kills the sign. And note the square: doubling $|\Delta x|$ does not double $U_s$, it quadruples it.
A $2.0$ kg ball sits on a shelf $1.5$ m above the floor. Take $g = 10\ \mathrm{m/s^2}$. Find $\Delta U_g$ as the ball is moved from the floor to the shelf, using two different choices for $y = 0$.
At the floor, $y = 0$. At the shelf, $y = 1.5$ m. So $\Delta U_g = m g\, \Delta y = (2.0)(10)(1.5) = 30$ J. Positive: potential energy increases as the ball goes up.
Now call the shelf $y = 0$. The floor is then at $y = -1.5$ m. Moving from floor to shelf is still $\Delta y = 0 - (-1.5) = +1.5$ m. So $\Delta U_g = (2.0)(10)(1.5) = 30$ J. Same answer.
The absolute $U_g$ at any height depends on where you put zero, but the change between two heights is the same for everyone. That change is what shows up in conservation problems.
One. Potential energy is a property of the system. Saying "the ball has $30$ J of potential energy" is shorthand for "the ball-Earth system has $30$ J of gravitational potential energy, relative to the chosen zero." When the ball falls, that energy does not leak out of the ball into the air. It converts into kinetic energy of the ball-Earth system. Earth is too massive to move noticeably, so almost all the kinetic energy lands in the ball. The bookkeeping always includes both partners, even when only one of them visibly moves.
Two. Elastic potential energy scales as the square of the displacement. Push a spring twice as far and you store four times the energy. Push three times as far and you store nine times. The everyday version of this surprise lives in cars: a vehicle at $60$ mph carries four times the kinetic energy of the same vehicle at $30$ mph. Square first, then scale.
Three. The zero of potential energy is a free choice. Every potential energy problem starts with an unstated question: where does $U = 0$ live? You answer it; the physics does not. A satellite orbiting a planet with $U_G = -1$ kJ at radius $r$ is in a real, physical state. The negative sign tells you that infinity was chosen as the zero, and bound orbits sit below it. Pick any other zero and the number changes, but the orbit does not.
This is the most natural mistake to make and the hardest to shake off. Gravitational potential energy sounds like a property the ball carries around with it, the way mass or charge are properties of the ball. Lifting the ball seems to fill it up with potential energy; dropping it seems to drain the potential energy back out. Everything seems to happen inside the ball.
But $mgh$ has Earth's gravity baked into it through $g$. Without a second interacting mass in the chosen system, there is no gravitational potential-energy term at all — even with the same ball at the same height.
Fix. Read $mgh$ as the potential energy of the ball-Earth system relative to a chosen ground. The number lives in the pair, not in the ball. The same applies to springs: $U_s = \tfrac{1}{2}k(\Delta x)^2$ belongs to the spring-block configuration, not to the spring or block alone.
This is the same linear-thinking trap that catches kinetic energy ("double the speed, double the KE"). The intuition is that effort scales with displacement, so twice as much push should mean twice as much storage.
But $U_s = \tfrac{1}{2} k (\Delta x)^2$ is quadratic in $\Delta x$. Doubling $|\Delta x|$ multiplies the energy by $2^2 = 4$. Tripling it multiplies by $9$.
Fix. Square first, then scale. Whenever a quantity has $(\Delta x)^2$ or $v^2$ in it, ask "what does the squared piece do?" before reaching for the multiplier. The factor of $\tfrac{1}{2}$ on the outside never changes the scaling; only the squared term does.
Two flavors of this. Flavor A: someone says "the floor is the real zero of potential energy," then gets confused when a different choice of zero produces negative values higher up. Flavor B: someone sees $U_G = -G m_1 m_2 / r$ on the formula sheet and concludes the formula must be wrong because energy cannot be negative.
Both errors come from thinking the zero of potential energy is set by nature, when it's something you pick. Negative $U$ means the system sits below whichever line you picked as zero. That's fine. A bound satellite always has $U_G < 0$ when the zero is at infinity, because moving the satellite to infinity would cost energy.
Fix. Negative values are fine. What matters physically is $\Delta U$ between configurations, which doesn't depend on where you put the zero. A negative $U_G$ tells you two things: someone chose $U_G = 0$ at $r = \infty$, and the system is bound (it can't escape on its own). Both are useful.
Ten scenarios. Five symbolic and factor-of-change, five numeric or conceptual. Pick one chip per part, hit Check, and your progress is saved in this browser as you go.