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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta h = 45 – 2 = 43\,\text{m}\] | The student’s center of mass drops from the bridge to the lowest point, so the vertical displacement is \(45\,\text{m}-2\,\text{m}=43\,\text{m}\). |
| 2 | \[x = \Delta h – L = 43 – 18 = 25\,\text{m}\] | The bungee cord (natural length \(L = 18\,\text{m}\)) does not stretch during the first \(18\,\text{m}\) of fall. The extra \(25\,\text{m}\) is the cord’s stretch \(x\). |
| 3 | \[m g \Delta h = \tfrac12 k x^2\] | Conservation of mechanical energy: the gravitational potential energy lost \(m g \Delta h\) equals the elastic potential energy \(\tfrac12 k x^2\) stored in the stretched cord at the lowest point (velocity momentarily zero). |
| 4 | \[k = \frac{2 m g \Delta h}{x^2}\] | Algebraically solve the energy equation for the spring constant \(k\). |
| 5 | \[k = \frac{2 (81)(9.8)(43)}{25^2}\] | Substitute \(m = 81\,\text{kg}\), \(g = 9.8\,\text{m/s}^2\), \(\Delta h = 43\,\text{m}\), and \(x = 25\,\text{m}\). |
| 6 | \[\boxed{k \approx 109.3\,\text{N/m}}\] | Evaluate the expression to find the required spring constant. |
Just ask: "Help me solve this problem."
An object is projected vertically upward from ground level. It rises to a maximum height [katex] H [/katex]. If air resistance is negligible, which of the following must be true for the object when it is at a height [katex] H/2 [/katex] ?
A runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A \(6 \, \text{kg}\) cube rests against a compressed spring with a force constant of \(1{,}800 \, \text{N/m}\), initially compressed by \(0.3 \, \text{m}\). Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of \(\mu_k = 0.12\) for \(3 \, \text{m}\), then ascends a \(12^\circ\) slope, stopping after \(4.5 \, \text{m}\). Determine the coefficient of kinetic friction on the slope.
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance \(h\) above the floor. A block of mass \(M\) is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance \(x\). The block is released and strikes the floor a horizontal distance \(D\) from the edge of the table. Air resistance is negligible. Derive expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
\(k \approx 109.3 \,\text{N/m}\)
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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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