AP Physics C: Mechanics

Unit 3

Work, Energy, and Power

8 practice questions · Free response

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

A force $F(x) = 3x^2$ N acts on an object from $x = 0$ to $x = 2$ m. How much work is done?
A potential energy function is $U(x) = 4x^2$ (J). What is the force at $x = 2$ m?
On a potential energy curve, a point of stable equilibrium is at a:
What is $\vec F \cdot \vec d$ for $\vec F = 3\hat i + 4\hat j$ N and $\vec d = 2\hat i - 1\hat j$ m?
How does escape speed depend on a planet's mass $M$ (same radius)?
A 1,000 kg car moves at 20 m/s while the engine exerts a 500 N forward force. What power does the engine deliver?
The gravitational potential energy of a satellite far from Earth:
A particle with total energy $E$ moves in a potential $U(x)$. It can't reach regions where:

Free response

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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.

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(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.

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(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: