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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]E_{\text{total}} = PE_{\text{max}} = KE_{\text{max}}[/katex] | Total mechanical energy is constant, equal to maximum PE (at highest point) and maximum KE (at lowest point). |
| 2 | [katex]E_{\text{total}} = KE + PE[/katex] | At any point, total mechanical energy is the sum of KE and PE. |
| 3 | [katex]PE = mgh[/katex] | Potential energy at any point, where [katex]h[/katex] is height above the lowest point. |
| 4 | [katex]mgh = \frac{1}{2} E_{\text{total}}[/katex] | When KE equals PE, each is half the total mechanical energy. |
| 5 | [katex]h = L(1 – \cos(\theta))[/katex] | Height in terms of the pendulum length [katex]L[/katex] and angle [katex]\theta[/katex]. |
| 6 | [katex]L(1 – \cos(\theta)) = \frac{1}{2} L[/katex] | Substitute [katex]h[/katex] into the equation from Step 4, using [katex]E_{\text{total}} = mgL[/katex] (max PE). |
| 7 | [katex]1 – \cos(\theta) = \frac{1}{2}[/katex] | Simplify the equation. |
| 8 | [katex]\cos(\theta) = \frac{1}{2}[/katex] | Rearrange the equation to solve for [katex]\cos(\theta)[/katex]. |
| 9 | [katex]\theta \approx 60^\circ[/katex] | Solve for [katex]\theta[/katex], knowing that [katex]\cos^{-1}\left(\frac{1}{2}\right) \approx 60^\circ[/katex]. |
Therefore, the angle [katex]\theta[/katex] at which the kinetic energy equals the potential energy is approximately [katex]\boxed{60^\circ}[/katex] from the VERTICAL. From the horizontal this angle would be 90°-60° = 30° which is less than 45°. Note the questions ask for the angle from the horizontal.
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A rocket of mass m is launched with kinetic energy K0, from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii in terms of the gravitational constant, G; the mass of the Earth, mE ; the radius of the Earth, RE ; and the mass of the rocket?
The efficiency of a pulley system is 55%. The
pulleys are used to raise a mass of 90.0 kg to a height of
5.60 m. What force is exerted on the rope of the pulley
system if the rope is pulled for 22 m in order to raise
the mass to the required height?
A uniform solid cylinder of mass [katex] M [/katex] and radius [katex] R [/katex] is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force [katex] F [/katex] , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle [katex] \theta [/katex], what is the kinetic energy of the cylinder in terms of [katex] F [/katex] and [katex] \theta [/katex]?
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
In 3.0 minutes, a ski lift raises 10 skiers at constant speed to a height of 85 m. The ski lift is 55° above the horizontal and the average mass of each skier is 67.5 kg. What is the average power provided by the tension in the cable pulling the lift?
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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