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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]T_{\text{max}} = mg[/katex] | The maximum tension [katex]T_{\text{max}}[/katex] that the wire can withstand is equal to the weight of the heaviest load it can support without breaking. Here, [katex]m = 70.0 \, \text{kg}[/katex] (mass of the person) and [katex]g = 9.8 \, \text{m/s}^2[/katex] (acceleration due to gravity). |
| 2 | [katex]T_{\text{max}} = 70.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2[/katex] | Calculate the maximum tension [katex]T_{\text{max}}[/katex] using the product of mass and acceleration due to gravity. |
| 3 | [katex]T_{\text{max}} = 686 \, \text{N}[/katex] | Performing the multiplication yields the maximum tension value. |
| 4 |
[katex]T = m_{\text{load}}(g + a)[/katex] |
Using netwons 2nd law, add all the forces acting on the load being lifted. In this case the Tension and weight act in opposite directions, thus [katex]T – m_{\text{load}}g = m_{\text{load}}a[/katex]. Re-arrange the equation for [katex]T[/katex] and factor out [katex]m_{\text{load}}[/katex] |
| 5 | [katex]686 \, \text{N} = 45.0 \, \text{kg} \cdot (9.8 \, \text{m/s}^2 + a)[/katex] | Since the wire can just barely support [katex]686 \, \text{N}[/katex], set the tension required to lift the load equal to this maximum. |
| 6 | [katex]9.8 \, \text{m/s}^2 + a = \frac{686 \, \text{N}}{45.0 \, \text{kg}}[/katex] | Divide both sides by the mass of the load to solve for the acceleration. |
| 7 | [katex]a = \frac{686 \, \text{N}}{45.0 \, \text{kg}} – 9.8 \, \text{m/s}^2[/katex] | Solve for [katex]a[/katex] by subtracting gravity’s acceleration from the result of the division. |
| 8 | [katex]a \approx 5.4 \, \text{m/s}^2[/katex] | Calculate the value of [katex]a[/katex]. This is the maximum vertical acceleration that can be achieved without breaking the wire. |
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A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?
A horizontal 300 N force pushes a 40 kg object across a horizontal 10 meter frictionless surface. After this, the block slides up a 20° incline. Assuming the incline has a coefficient of kinetic friction of 0.4, how far along the incline with the object slide?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
In the diagram above block A has a mass of 3.2 kg and block B a mass of 2.4 kg. The pulley is frictionless and has no mass.
Two students push a 1750 kg car with a force of 758 N along a perfectly level road at a constant velocity of 4.00 m/s. Find the force of friction.
[katex]a \approx 5.4 \, \text{m/s}^2[/katex]
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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