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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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In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
A \(25 \, \text{g}\) steel ball is attached to the top of a \(24 \, \text{cm}\)-diameter vertical wheel of negligible mass. Starting from rest, the wheel accelerates at \(470 \, \text{rad/s}^2\). The ball is released after \(\frac{3}{4}\) of a revolution. How high does it go above the center of the wheel?
A block with a mass of \( 4 \) \( \text{kg} \) is attached to a spring on the wall that oscillates back and forth with a frequency of \( 4 \) \( \text{Hz} \) and an amplitude of \( 3 \) \( \text{m} \). What would the frequency be if the block were replaced by one with one‑fourth the mass and the amplitude of the block is increased to \( 9 \) \( \text{m} \)?
A mass–spring system is oscillating in simple harmonic motion. At the exact moment the mass passes through its equilibrium position, which of the following statements is true?
A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?
A \(100 \, \text{kg}\) person is riding a \(10 \, \text{kg}\) bicycle up a \(25^\circ\) hill. The hill is long and the coefficient of static friction is \(0.9\). The person rides \(10 \, \text{m}\) up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of \(7 \, \text{m}\) before squeezing hard on the brakes locking the wheels, how much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is \(0.65\)?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
A small block moving with a constant speed \(v\) collides inelastically with a block \(M\) attached to one end of a spring \(k\). The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A rubber ball bounces off of a wall with an initial speed \(v\) and reverses its direction so its speed is \(v\) right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
Students attach a thin strip of metal to a table so that the strip is horizontal in relation to the ground. A section of the strip hangs off the edge of the table. A mass is secured to the end of the hanging section of the strip and is then displaced so that the mass-strip system oscillates, as shown in the figure. Students make various measurements of the net force F exerted on the mass as a result of the force due to gravity and the normal force from the strip, the vertical position y of the mass above and below its equilibrium position y. and the period of oscillation T’ when the mass is displaced by different amplitudes A. Which of the following explanations is correct about the evidence required to conclude that the mass undergoes simple harmonic motion?
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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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