| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( F_{\text{net}} = m_{4}a \) | The net force acting on the \(4 \, \text{kg}\) block is equal to its mass times its acceleration. |
| 2 | \( F_{\text{gravity}} = m_{4}g \) | The force of gravity acting on the \(4 \, \text{kg}\) block is equal to its mass times the acceleration due to gravity \(g\). |
| 3 | \( F_{\text{net}} = m_{4}g – T \) | The net force acting on the \(4 \, \text{kg}\) block is the gravitational force minus the tension in the string. |
| 4 | \( m_{4}a = m_{4}g – T \) | Substitute the net force into Newton’s second law. |
| 5 | \( T = m_{1}g + m_{1}a + m_{2}g + m_{2}a \) | The tension in the string also depends on the forces acting on the \(1.0\, \text{kg}\) and \(2.0 \, \text{kg}\) blocks. |
| 6 | \( T = m_{1}(g + a) + m_{2}(g + a) \) | Combine the tensions for the \(1.0 \, \text{kg}\) and \(2.0 \, \text{kg}\) blocks since they share the same strings. |
| 7 | \( m_{4}a = m_{4}g – \left[m_{1}(g + a) + m_{2}(g + a)\right] \) | Substitute the tension \(T\) from step 6 into the equation from step 4. |
| 8 | \( 4a = 4g – (1 + 2)(g + a) \) | Substitute \(m_{4} = 4\, \text{kg}\), \(m_{1} = 1\, \text{kg}\), and \(m_{2} = 2\, \text{kg}\). |
| 9 | \( 4a = 4g – 3(g + a) \) | Combine the masses for the \(1.0 \, \text{kg}\) and \(2.0 \, \text{kg}\) blocks. |
| 10 | \( 4a = 4g – 3g -3a \) | Distribute the 3 to both terms. |
| 11 | \( 7a = g \) | Combine like terms to isolate \(a\). |
| 12 | \( a = \frac{g}{7} \approx \frac{9.8\, \text{m/s}^2}{7} \approx 1.4\, \text{m/s}^2 \) | Solve for \(a\), the acceleration of the \(4 \, \text{kg}\) block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(T = m_4 g – m_4 a\) | The tension in the string is the gravitational force on the \(4 \, \text{kg}\) block minus the force due to its acceleration. |
| 2 | \(T = 4 \times 9.8 – 4 \times 1.4\) | Substitute \(m_4 = 4\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\), and \(a = 1.4\, \text{m/s}^2\). |
| 3 | \(T = 39.2 – 5.6\) | Calculate the products. |
| 4 | \( T = 33.6 \, \text{N} \) | Final value for the tension in the string supporting the \(4 \, \text{kg}\) block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_1 = m_1 (g + a) \) | The tension in the string is the gravitational force on the \(1 \, \text{kg}\) block plus the force due to its acceleration. |
| 2 | \( T_1 = 1 \times (9.8 + 1.4) \) | Substitute \(m_1 = 1\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\), and \(a = 1.4\, \text{m/s}^2\). |
| 3 | \( T_1 = 1 \times 11.2 \) | Combine the terms inside the parentheses. |
| 4 | \( T_1 = 11.2 \, \text{N} \) | Final value for the tension in the string connected to the \(1 \, \text{kg}\) block. |
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A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
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?
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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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