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| Derivation or Formula | Reasoning |
|---|---|
| \[v_{\max}=\omega A\] | In simple harmonic motion, the maximum speed occurs at equilibrium \((x=0)\) and equals \(v_{\max}=\omega A\). |
| \[v=0.50\,v_{\max}=0.50\,\omega A\] | The problem states the speed is \(50\%\) of its maximum value, so \(v\) is half of \(v_{\max}\). |
| \[v^2=\omega^2(A^2-x^2)\] | This SHM relation connects speed \(v\) and position \(x\) using energy (or standard SHM kinematics) without calculus. |
| \[(0.50\,\omega A)^2=\omega^2(A^2-x^2)\] | Substitute \(v=0.50\,\omega A\) into \(v^2=\omega^2(A^2-x^2)\). |
| \[0.25\,\omega^2A^2=\omega^2(A^2-x^2)\] | Square the left side and simplify. |
| \[0.25A^2=A^2-x^2\] | Cancel \(\omega^2\) from both sides (nonzero for SHM). |
| \[x^2=A^2-0.25A^2=0.75A^2=\frac{3}{4}A^2\] | Solve algebraically for \(x^2\). |
| \[|x|=\sqrt{\frac{3}{4}}A=\frac{\sqrt{3}}{2}A\] | The question asks for the magnitude of position, so take the positive square root: \(|x|\). |
| \[\boxed{|x|=\frac{\sqrt{3}}{2}A}\] | Final answer in terms of only \(A\). |
| Derivation or Formula | Reasoning |
|---|---|
| \[E_{\text{tot}}=\frac{1}{2}kA^2\] | Total mechanical energy in SHM equals the maximum spring potential energy, which occurs at \(|x|=A\). |
| \[U_s=\frac{1}{2}kx^2\] | The elastic (spring) potential energy at position \(x\) is \(U_s=\frac{1}{2}kx^2\). |
| \[U_s=0.50\,E_{\text{tot}}\] | The problem states the spring potential energy is \(50\%\) of the total energy. |
| \[\frac{1}{2}kx^2=0.50\left(\frac{1}{2}kA^2\right)\] | Substitute the expressions for \(U_s\) and \(E_{\text{tot}}\). |
| \[\frac{1}{2}kx^2=\frac{1}{4}kA^2\] | Simplify the right-hand side: \(0.50\cdot\frac{1}{2}=\frac{1}{4}\). |
| \[x^2=\frac{1}{2}A^2\] | Cancel \(k\) and multiply both sides by \(2\) to isolate \(x^2\). |
| \[|x|=\sqrt{\frac{1}{2}}A=\frac{A}{\sqrt{2}}\] | Take the square root and use magnitude \(|x|\) since the oscillator can be at \(+x\) or \(-x\). |
| \[\boxed{|x|=\frac{A}{\sqrt{2}}}\] | Final answer in terms of only \(A\). |
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A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
A box having a mass of \( 1.5 \) \( \text{kg} \) is accelerated across a table at \( 1.5 \) \( \text{m/s}^2 \). The coefficient of kinetic friction on the box is \( 0.3 \).
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.
A pendulum has a period of \(2.0 \, \text{s}\) on Earth. What is its length?
A force \(F\) is exerted by a broom handle on the head of a broom, which has a mass \(m\). The handle is at an angle \(\theta\) to the horizontal. The work done by the force on the head of the broom as it moves a distance \(d\) across a horizontal floor is
A ski lift carries skiers along a \(695 \, \text{m}\) slope inclined at \(34^\circ\). To lift a single rider, it is necessary to move \(72 \, \text{kg}\) of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If \(65\%\) of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
A theme park ride consists of a large vertical wheel of radius \( R \) that rotates counterclockwise on a horizontal axle through its center. The cars on the wheel move at a constant speed \( v \). Points \( A \) and \( D \) represent the position of a car at the highest and lowest point of the ride, respectively. While passing point \( A \), a student releases a small rock of mass \( m \), which falls to the ground without hitting anything. Which of the following best represents the kinetic energy of the rock when it is at the same height as point \( D \)?
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
An apple is released from rest \(500 \, \text{m}\) above the ground. Due to the combined forces of air resistance and gravity, it has a speed of \(40 \, \text{m/s}\) when it reaches the ground. What percentage of the initial mechanical energy of the apple–Earth system was dissipated due to air resistance? Take the potential energy of the apple–Earth system to be zero when the apple reaches the ground.
\(|x|=\frac{\sqrt{3}}{2}A\), \(|x|=\frac{A}{\sqrt{2}}\)
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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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