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| 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. |
Just ask: "Help me solve this problem."
Shown above are three masses of \(6 \, \text{kg}\), \(3 \, \text{kg}\), and \(1 \, \text{kg}\) (in order from left to right). You pull on the 1kg mass with a force \(F\) of \(15 \, \text{N}\) along a frictionless surface.
A stone hangs by a fine thread from the ceiling, and a section of the same thread dangles from the bottom of the stone. If a person gives a sharp pull on the dangling thread, where is the thread likely to break: below the stone or above it? What if the person gives a slow and steady pull?
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of \( 38.0^\circ \) with respect to the horizontal. Suppose a wooden crate with a mass of \( 25.0 \) \( \text{kg} \) is placed on Baldwin Street. An additional force of \( 59 \) \( \text{N} \) must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is \( 0.599 \), what is the magnitude of the frictional force?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end, and the other endpoint is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the magnitude and direction of the force between the cable and the wall and of the force between the steel beam and the wall.
A spring is connected to a wall and a horizontal force of \( 80.0 \) \( \text{N} \) is applied. It stretches \( 25 \) \( \text{cm} \); what is its spring constant?
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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] |
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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