Revision sheet: work, energy and power (PHYS 101)

On your formula sheet: this chapter's lines, and when each one holds

• • $W=F\cdot d$: A dot product: $W=Fd\cos\theta$ for a constant force along a straight displacement. A force at $90^{\circ}$ to the motion does no work; kinetic friction does negative work.
• • $KE=\frac{1}{2}mv^{2}$: Built on the speed, so it never depends on the direction and is never negative.
• • $W_{net}=\Delta KE$: $W_{net}$ is the work of ALL the forces, gravity and springs included.
• • $W_{non-cons.}=\Delta KE+\Delta PE$: Gravity and springs now sit inside $\Delta PE$: counting them again in $W_{non-cons.}$ counts them twice. Friction makes $W_{non-cons.}$ negative.
• • $PE_{grav}=mgh$: Near the Earth's surface. $h$ is measured from a level you choose, and only differences of $PE$ ever matter.
• • $PE_{spring}=\frac{1}{2}kx^{2}$: $x$ is the stretch or the compression measured from the natural length, not the position of the object.
• • $F=-kx$: The spring pulls or pushes back toward its natural length: the minus sign is a direction, the magnitude is $kx$.
• • $power=energy/time=W/t=F\cdot v$: $\frac{W}{t}$ is an average power; $F\cdot v$ is the power at the instant the speed is $v$. In watts.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam. This sheet writes $U$ for potential energy where the formula sheet writes $PE$: same quantity.

The three energies and the one line that uses them

• • Work of a constant force: $W = Fd\cos\theta$, with $\theta$ the angle between the FORCE and the DISPLACEMENT, not the angle with anything else on the diagram. The formula sheet writes it as a dot product, $W = \vec{F}\cdot\vec{d}$, which is the same line.
• • Kinetic energy $E_{k} = \frac{1}{2}mv^{2}$, always positive, and proportional to $v^{2}$: double the speed, quadruple the energy.
• • Gravitational potential energy $U = mg\Delta y$ with $g = 9.8$ m/s$^{2}$, where $\Delta y$ is the VERTICAL height above a level you choose and write down. Elastic: $U = \frac{1}{2}kx^{2}$.
• • The only equation you need: $E_{k,i} + U_{i} + W_{\text{nc}} = E_{k,f} + U_{f}$, where $W_{\text{nc}}$ is the work of friction and of any applied force, negative for friction.
• • Power $P = \dfrac{W}{t} = Fv$ in watts, with $t$ in SECONDS. Efficiency $= \dfrac{\text{useful output}}{\text{total input}}$, never above 1. At constant force the power follows the speed: pushed from rest by a fixed force, a vehicle takes a power that grows in a straight line from zero.

Everything in this chapter is one of two questions: what is the energy HERE, and what happened to it between here and there. If your page has more than one line of algebra per state, you are solving it as a forces problem in disguise.

Work: the angle, the sign, and the forces that do none

• • A force at less than $90^{\circ}$ to the motion does POSITIVE work and adds energy. At exactly $90^{\circ}$ it does NO work. Beyond $90^{\circ}$ it does NEGATIVE work and removes energy.
• • The three forces that almost never do work: the normal force on any surface, the tension in a pendulum string, and any centripetal force on a circular path. All three are perpendicular to the velocity at every instant.
• • Friction on a sliding object is always at $180^{\circ}$ to the motion: $W_{f} = -fd$, with the minus sign written, not remembered.
• • A force that varies with position has no single value to put in $Fd$: its work is the AREA between the graph of $F$ against position and the horizontal axis.

Bar charts: an energy budget you can check

• • One bar chart per state: $U$, $E_{k}$, and, once friction has acted, the energy already turned into heat, $f\,d$.
• • Without friction the total height of the bars never changes. With friction the heat bar grows by $f\,d$, and the mechanical bars shrink by exactly as much.
• • A spring adds a third stored energy, $\frac{1}{2}kx^{2}$, which turns into $E_{k}$ as the spring relaxes.

If the bars of the final state are taller than those of the first, an applied force has done positive work, or a sign has been lost.

Answer in letters first, then three checks

• • Write the answer in letters before any number, in terms of the data the statement names. The numbers go in once, at the very end, and the letters are what the three checks below can test.
• • Parcel stopped by a buffer spring at the foot of a rough chute, from rest to the greatest compression: $\frac{1}{2}kx^{2} = F(D + x)$ with $F = mg(\sin\theta - \mu_k\cos\theta)$, a quadratic in $x$, so $x = \dfrac{F + \sqrt{F^{2} + 2kFD}}{k}$.
• • Check 1, the unit: $F/k$ is N over N/m, a length, and the square root is in newtons, so $x$ is in metres.
• • Check 2, a limit case: with $D = 0$, $x = 2F/k$, twice the compression at which the spring would simply hold the parcel.
• • Check 3, the sense of variation: a stickier chute must give a smaller $x$. With $8.0$ kg, $30^{\circ}$, $D = 1.50$ m and $k = 900$ N/m: $x = 0.408$ m without friction, $0.322$ m with $\mu_k = 0.20$.

A literal answer that fails one of the three checks is wrong before any number is computed, and the failed check says where: a wrong unit is a lost $k$ or $g$, a wrong sense of variation is a sign on the work of friction.