Notes
Work as an integral
- $\displaystyle W = \int \vec F \cdot d\vec r$. In one dimension, $\displaystyle W = \int_{x_1}^{x_2} F(x)\,dx$, the area under the $F$ vs $x$ graph.
- Dot product: $\vec A \cdot \vec B = A_xB_x + A_yB_y = AB\cos\theta$.
- Work-energy theorem: $W_{net} = \Delta K$.
- Spring: $\displaystyle W_{spring} = -\frac{1}{2}kx^2$ from equilibrium, so $\displaystyle U_s = \frac{1}{2}kx^2$.
Potential energy and force
- For a conservative force, $\displaystyle \Delta U = -\int F\,dx$ and $\displaystyle F = -\frac{dU}{dx}$.
- On a potential energy curve $U(x)$, the force points "downhill." Stable equilibrium is at a minimum; unstable at a maximum. The particle can't go where $U > E$ (turning points).
- Gravitational potential energy (anywhere, not just near Earth): $\displaystyle U_G = -\frac{Gm_1m_2}{r}$.
- Escape speed: $\displaystyle v_{esc} = \sqrt{\frac{2GM}{R}}$.
Conservation of energy
$K_i + U_i + W_{nc} = K_f + U_f$, where $W_{nc}$ is work by nonconservative forces like friction.
Power
$\displaystyle P = \frac{dW}{dt} = \vec F \cdot \vec v$.
Practice
$\displaystyle \int_0^2 3x^2\,dx = 8$ J.
$\displaystyle F = -\frac{dU}{dx} = -8x$.
Small displacements create a force back toward the minimum.
$3(2) + 4(-1) = 2$ J.
$\displaystyle v_{esc} = \sqrt{\frac{2GM}{R}}$.
$P = Fv = 500(20) = 10{,}000$ W.
$\displaystyle U = -\frac{GMm}{r}$ rises toward 0 as $r$ grows.
Kinetic energy can't be negative.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 2.0 kg object moves along the $x$-axis under a single force $F(x) = 12 - 4x$ (N). It starts at $x = 0$ with speed 1.0 m/s.
(a) Find the work done by the force as the object moves from $x = 0$ to $x = 3$ m.
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$\displaystyle W = \int_0^3 (12 - 4x)\,dx = \left[12x - 2x^2\right]_0^3 = 36 - 18 = 18$ J.
Check off the points you earned:
(b) Find the object's speed at $x = 3$ m.
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$\displaystyle \frac{1}{2}(2.0)v^2 = \frac{1}{2}(2.0)(1.0)^2 + 18 = 19 \Rightarrow v = \sqrt{19} \approx 4.4$ m/s.
Check off the points you earned:
(c) Find a potential energy function $U(x)$ for this force with $U(0) = 0$.
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$\displaystyle U(x) = -\int_0^x (12 - 4x)\,dx = -12x + 2x^2$ (J).
Check off the points you earned: