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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. |
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A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
A \(20 \, \text{g}\) piece of clay moving at a speed of \(50 \, \text{m/s}\) strikes a \(500 \, \text{g}\) pendulum bob at rest. The length of a string is \(0.8 \, \text{m}\). After the collision, the clay-bob system starts to oscillate as a simple pendulum.
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
A 0.5 kg pendulum bob is raised to 1.0 m above the floor, as shown in the figure. The bob is then released from rest. When the bob is 0.8 m above the floor, its speed is most nearly
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude \( A \) on a horizontal frictionless surface. Determine the following in terms of only the variable \( A \).
A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
A block of mass \( m \) is attached to a horizontal spring with spring constant \( k \) and undergoes simple harmonic motion with amplitude \( A \) along the \( x \)-axis. Which of the following equations could represent the position \( x \) of the object as a function of time?
A ski tow carries people to the top of a nearby mountain. It operates on a slope of angle \( 15.7^\circ \) of length \( 260 \) \( \text{m} \). The rope moves at a speed of \( 13.0 \) \( \text{km/h} \) and provides power for \( 54 \) riders at one time, with an average mass per rider of \( 67.0 \) \( \text{kg} \).
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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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