| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Choose }a: \text{4 kg down, others up}\] | Set a common magnitude \(a\). The 4 kg block moves downward; 2 kg and 1 kg move upward. |
| 2 | \[m_1 g- T_1 = m_1 a\] | Newton’s 2ⁿᵈ for 4 kg (down is positive). |
| 3 | \[T_1- m_2 g- T_2 = m_2 a\] | 2 kg block: up is positive. |
| 4 | \[T_2- m_3 g = m_3 a\] | 1 kg block: up is positive. |
| 5 | \[T_1 = m_1(g-a)\qquad T_2 = m_3(g+a)\] | Solve Steps 2 and 4 for \(T_1\) and \(T_2\). |
| 6 | \[m_1(g-a)-m_2 g-m_3(g+a)=m_2 a\] | Substitute \(T_1\) and \(T_2\) into Step 3. |
| 7 | \[a=g\frac{m_1-m_2-m_3}{m_1+m_2+m_3}\] | Collect like terms and isolate \(a\). |
| 8 | \[a=9.8\,(\mathrm{m/s^2})\frac{4-2-1}{4+2+1}=1.4\,\mathrm{m/s^2}\] | Insert \(m_1=4\,\text{kg},\; m_2=2\,\text{kg},\; m_3=1\,\text{kg}\). |
| 9 | \[\boxed{a=1.4\,\mathrm{m/s^2}\text{ (4 kg downward)}}\] | Final acceleration magnitude and direction for the 4 kg block. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T_1=m_1(g-a)\] | From Step 5 of part (a). |
| 2 | \[T_1=4\,(9.8-1.4)=33.6\,\mathrm{N}\] | Insert \(m_1=4\,\text{kg}\) and \(a=1.4\,\mathrm{m/s^2}\). |
| 3 | \[\boxed{T_1=33.6\,\mathrm{N}}\] | Numerical tension in the string holding the 4 kg block. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T_2=m_3(g+a)\] | From Step 5 of part (a). |
| 2 | \[T_2=1\,(9.8+1.4)=11.2\,\mathrm{N}\] | Insert \(m_3=1\,\text{kg}\) and \(a=1.4\,\mathrm{m/s^2}\). |
| 3 | \[\boxed{T_2=11.2\,\mathrm{N}}\] | Numerical tension in the lower string. |
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A traffic light hangs from a pole as shown in the diagram. The uniform aluminum pole AB is of length \( 7.20 \) \( \text{m} \) and has a mass of \( 12.0 \) \( \text{kg} \). The mass of the traffic light is \( 21.5 \) \( \text{kg} \). The point C is located \( 3.80 \) \( \text{m} \) vertically above the pivot A. A massless horizontal cable CD is attached at C and connects to the pole at point D, where the pole makes an angle of \( 37^{\circ} \) with the cable.
What is the weight of a person who has a mass of \(75 \, \text{kg}\)?
A \(2.2 \times 10^{21} \, \text{kg}\) moon orbits a distant planet in a circular orbit of radius \(1.5 \times 10^8 \, \text{m}\). It experiences a \(1.1 \times 10^{19} \, \text{N}\) gravitational pull from the planet. What is the moon’s orbital period in Earth days?
Are astronauts really “weightless” while in orbit?
A spring with spring constant \( k = 2.3 \) \( \text{N/m} \) is attached to an object of mass \( 10 \) \( \text{kg} \). If the object is hung from the ceiling by this spring, how much would the spring be stretched?
When the brakes of an automobile are applied, the road exerts the greatest retarding force
A car suddenly stops and a passenger lurches forward. This motion is best explained by Newton’s ____ Law.
A \( 15 \) \( \text{N} \) force is pushing a \( 40 \) \( \text{N} \) block down a incline. The angle of the inline is \( \alpha = 40^{\circ} \). The coefficient of static friction between the block and the incline is \( \mu_s = 0.75 \) and the coefficient of kinetic friction is \( \mu_k = 0.65 \).
A skier with a mass of \(58 \, \text{kg}\) glides up a snowy incline that forms an angle of \(28^\circ\) with the horizontal. The skier initially moves at a speed of \(7.2 \, \text{m/s}\). After traveling a distance of \(2.3 \, \text{m}\) up the slope, the skier’s speed reduces to \(3.8 \, \text{m/s}\).
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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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