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| Derivation/Formula | Reasoning |
|---|---|
| \[ \omega = \sqrt{\frac{k}{m}} \] | This is the formula for the angular frequency of a mass-spring system, where \(k=20.0\,\text{N/m}\) and \(m=1.5\,\text{kg}\). |
| \[ \omega = \sqrt{\frac{20.0}{1.5}} \approx 3.65\,\text{rad/s} \] | Substitute the given values to calculate \(\omega\). |
| \[ f = \frac{\omega}{2\pi} \approx \frac{3.65}{6.28} \approx 0.582\,\text{Hz} \] | Convert the angular frequency to the ordinary frequency using \(f=\omega/(2\pi)\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[ v_{\text{max}} = A\,\omega \] | The maximum speed in simple harmonic motion is the product of the amplitude \(A\) and the angular frequency \(\omega\). |
| \[ v_{\text{max}} = 0.10\,\text{m} \times 3.65\,\text{rad/s} \approx 0.365\,\text{m/s} \] | Substitute the amplitude \(A=0.10\,\text{m}\) and the computed \(\omega\) into the formula. |
| \[ \text{Occurs at } x=0 \] | The maximum speed occurs at the equilibrium position where the displacement is zero. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ a_{\text{max}} = \omega^2\,A \] | The maximum acceleration in simple harmonic motion is given by \(a_{\text{max}}=\omega^2 A\). |
| \[ a_{\text{max}} = (3.65\,\text{rad/s})^2 \times 0.10\,\text{m} \approx 1.33\,\text{m/s}^2 \] | Substitute \(\omega \approx 3.65\,\text{rad/s}\) and \(A = 0.10\,\text{m}\) into the equation. |
| \[ \text{Occurs at } x = \pm 0.10\,\text{m} \] | The magnitude of acceleration is maximum at the extreme positions (\(x=\pm A\)) of the oscillation. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ E = \frac{1}{2}\,k\,A^2 \] | The total mechanical energy in a mass-spring system is stored as potential energy in the spring at maximum displacement. |
| \[ E = \frac{1}{2} \times 20.0\,\text{N/m} \times (0.10\,\text{m})^2 \] | Substitute the given values \(k=20.0\,\text{N/m}\) and \(A=0.10\,\text{m}\) into the energy formula. |
| \[ E = 0.1\,\text{J} \] | Simplify the expression to obtain the total energy of the system. |
| Derivation/Formula | Reasoning |
|---|---|
| \[ x(t) = A\,\cos(\omega t + \phi) \] | This is the general solution for the displacement in simple harmonic motion, where \(\phi\) is the phase constant. |
| \[ x(0) = A\,\cos(\phi) = 0.10\,\text{m} \] | At \(t=0\), the mass is released from rest at \(x=0.10\,\text{m}\), which implies \(\phi = 0\) because \(\cos(0)=1\). |
| \[ x(t) = 0.10\,\text{m}\,\cos\Big(\sqrt{\frac{20.0}{1.5}}\,t\Big) \] | Substitute \(A=0.10\,\text{m}\), \(\omega=\sqrt{\frac{20.0}{1.5}}\), and \(\phi=0\) into the general solution to obtain the displacement as a function of time. |
Just ask: "Help me solve this problem."
An object at rest suddenly explodes into two fragments (m1 and m2) by an explosion. Fragment m1 acquires 3 times the kinetic energy of the other. What is the ratio of m1 to m2?
The box in the diagram is sliding to the right across a horizontal table, under the influence of the forces shown. Which force(s) is doing negative work on the box? Solve this both mathematically and justify using conceptual reasoning of what negative work really means.
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?
A bullet (mass: \(0.05 \, \text{kg}\)) is fired horizontally (\(v = 200 \, \text{m/s}\)) at a block (mass: \(1.3 \, \text{kg}\)) initially at rest on a frictionless surface. The block is attached to a spring (\(k = 2500 \, \text{N/m}\)). The bullet becomes embedded. Calculate:
In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
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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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