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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{spring, final}}[/katex] | Apply the law of conservation of energy. Initially, the climber has potential energy due to gravity, which gets converted into kinetic energy and the potential energy of the spring (the rope) when the climber is momentarily at rest. |
| 2 | [katex]mg(h + x) = \frac{1}{2}kx^2[/katex] | Express the potential energy due to height ([katex]mgh[/katex]) and potential energy stored in the stretched spring ([katex]\frac{1}{2}kx^2[/katex]), where [katex]m[/katex] is mass, [katex]g[/katex] is acceleration due to gravity (9.8 m/s2), [katex]h[/katex] is the original fall distance, and [katex]x[/katex] is the additional stretch in the rope. |
| 3 | [katex]84.4 \times 9.8 \times (0.627 + x) = \frac{1}{2} \times 1340 \times x^2[/katex] | Substitute the values [katex]m = 84.4\,kg[/katex], [katex]g = 9.8\,m/s^2[/katex], [katex]k = 1340\,N/m[/katex], and [katex]h = 0.627\,m[/katex] into the energy equation. |
| 4 | Solve for [katex]x[/katex] | Simplify and rearrange the equation introduced in Step 3 to form a quadratic equation in the standard form [katex]ax^2 + bx + c = 0[/katex]. Solve the quadratic using the formula [katex]x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/katex]. |
| 5 | [katex]670x^2 – 825.712x – 515.7308 = 0[/katex] | Perform the calculation to get the quadratic equation: [katex]670x^2 – 825.712x – 515.7308 = 0[/katex]. |
| 7 | [katex]x \approx 1.688\,m[/katex] or [katex]x \approx -0.456\,m[/katex] | Calculate the roots, by graphing or using the quadratic formula. Note the negative root (approximately -0.456 m) is not physically meaningful since stretch can’t be negative. |
| 8 | [katex]x \approx \boxed{ 1.688\,m} [/katex] | The feasible physical solution is that the rope stretches about 1.688 meters when it brings the climber to rest. |
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A mechanic pushes a [katex]2500 \, \text{kg}[/katex] car from rest to a final speed [katex]v[/katex] by doing [katex]5.0 \times 10^3 \, \text{J}[/katex] of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
A proton (mp = 1.67 x10-27 kg) is being accelerated along a straight line at 3.6 ×1015 m/s2 in a machine. The proton has an initial speed of 2.4 x107 m/s and travels 3.5 cm.
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
In 3.0 minutes, a ski lift raises 10 skiers at constant speed to a height of 85 m. The ski lift is 55° above the horizontal and the average mass of each skier is 67.5 kg. What is the average power provided by the tension in the cable pulling the lift?
A 81 kg student dives off a 45 m tall bridge with an 18 m long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be 2.0 m above the water?
1.688 m
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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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