| Derivation or Formula | Reasoning |
|---|---|
| \[\sum F_x = 0 \;\Rightarrow\; a_{\text{ground}} = 0\] | With the wagon just beginning to accelerate, no horizontal force acts on the ball, so, by Newton\’s first law, its horizontal acceleration relative to the ground is \(0\). |
| \[v_{\text{ground}} = v_i = 0\] | The ball and wagon were initially at rest; hence the ball keeps the same ground speed (zero) when the wagon starts to speed up. |
| \[\vec F_{\text{pseudo}} = -m\,\vec a_w\] | In the wagon\’s accelerating frame, a backward pseudo-force appears, equal and opposite to the wagon\’s forward acceleration \(\vec a_w\). |
| \[\vec a_{\text{wagon}} = -\vec a_w\] | That pseudo-force gives the ball an apparent backward acceleration of magnitude \(a_w\) relative to the wagon. |
| \[\Delta x_{\text{wagon}} = -\tfrac12 a_w t^2\] | With zero initial relative velocity in the wagon, the ball\’s displacement in the wagon frame grows backward (negative) quadratically in time. |
| \[\text{Path}_{\text{ground}} : \text{straight line, constant position}\] | Seen from the ground, the ball simply “hangs in space” while the wagon pulls forward beneath it. |
| \[\text{Path}_{\text{wagon}} : \text{moves to back wall}\] | Seen from inside the wagon, the ball drifts straight toward the rear, striking the back end if the acceleration continues. |
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Which of the following must be true for an object at translational equilibrium?
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?
A box is sliding down an incline at a constant speed of \( 2 \, \text{m/s} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \, \text{N} \). Derive an equation for the force of gravity acting on the box.
A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?
An elastic cord is \( 80\) \( \text{cm} \) long when it is supporting a mass of \( 10. \) \( \text{kg} \) hanging from it at rest. When an additional \( 4.0 \) \( \text{kg} \) is added, the cord is \( 82.5 \) \( \text{cm} \) long.
Find the tension in each cable supporting the gymnast who weighs \( 600 \) \( \text{N} \). The gymnast is at rest, holding a junction point where two cables are attached: one cable is horizontal, and the second cable is attached to the ceiling making an angle of \( 37^{\circ} \) above the horizontal, as shown in the diagram.
A car suddenly stops and a passenger lurches forward. This motion is best explained by Newton’s ____ Law.
A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.
Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
\(\text{Relative to ground: ball remains at its initial position while the wagon moves forward}\)
\(\text{Relative to wagon: ball accelerates backward with } a = -a_w\)
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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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