| Derivation/Formula | Reasoning |
|---|---|
| \[F = k \Delta x\] | Hooke’s law relates the spring force \(F\) to the spring constant \(k\) and the compression \(\Delta x\). |
| \[k \Delta x = mg\] | At equilibrium, the upward spring force balances the weight \(mg\) of the girl. |
| \[k = \frac{mg}{\Delta x}\] | Rearrange algebraically to solve for the spring constant \(k\). |
| \[k = \frac{(40)(9.8)}{0.12} = 3.3 \times 10^3\, \text{N/m}\] | Substitute \(m = 40\,\text{kg}\), \(g = 9.8\,\text{m/s}^2\), and \(\Delta x = 0.12\,\text{m}\), then compute. |
| \[\boxed{k = 3.3 \times 10^3\, \text{N/m}}\] | Final value of the spring constant. |
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A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
Which of the following statements about the acceleration due to gravity is TRUE?
The distance from earth to sun is \(1 \, \text{AU}\). The distance from Saturn to sun is \(9 \, \text{AU}\). Find the period of Saturn’s orbit in years. You can assume that the orbits are circular.
An object has a mass of \(10 \, \text{kg}\). For each case below answer the questions and provide an example.
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
An object weighs \( 432 \) \( \text{N} \) on the surface of Earth. At a height of \( 3R_{\text{Earth}} \) above Earth’s surface, what is its weight?
Two wires support an unknown mass as shown in the diagram. The tension in the left wire is measured to be \( 17.5 \) \( \text{N} \) and the tension in the right wire is \( 30.3 \) \( \text{N} \). The left wire makes an angle of \( 30^{\circ} \) with the horizontal, and the right wire makes an angle of \( 60^{\circ} \) with the horizontal. What is the mass of the object?
Why is the stopping distance of a truck much shorter than for a train going the same speed? Hint: try deriving a formula or stopping distance.
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
The exoplanet HD 69830b has a mass 10 times that of the Earth and a radius 5 times that of the Earth. The value of g on HD 69830b is most nearly
\(k = 3.3 \times 10^{3}\,\text{N/m}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| 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] |
Metric Prefixes
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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