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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[PE_{\text{top}} = mgh\] | Calculate the potential energy at the top of the incline. Here, the potential energy is given by the product of mass \(m\), gravitational acceleration \(g\), and height \(h\) of the incline. |
| 2 | \[KE_{\text{bottom}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\] | Calculate the kinetic energy at the bottom, considering both translational and rotational motion. \(I\) is the moment of inertia and \(\omega\) is the angular velocity. |
| 3 | \[I = \frac{2}{5}mR^2\] | Use the moment of inertia formula for a solid sphere. |
| 4 | \[\omega = \frac{v}{R}\] | Relate the angular velocity to translational velocity using \(\omega = \frac{v}{R}\) for rolling without slipping. |
| 5 | \[\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 = mgh\] | Substitute \(I\) and \(\omega\) into the kinetic energy equation and equate it to potential energy using conservation of energy. |
| 6 | \[\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh\] | Simplify the equation by calculating \(\omega^2 = \frac{v^2}{R^2}\). |
| 7 | \[\frac{7}{10}mv^2 = mgh\] | Combine the terms \(\frac{1}{2}mv^2\) and \(\frac{1}{5}mv^2\). |
| 8 | \[v^2 = \frac{10}{7}gh\] | Cancel the mass \(m\) from both sides and solve for \(v^2\). |
| 9 | \[v = \sqrt{\frac{10}{7} \times 9.8 \times 1.0}\] | Substitute \(g = 9.8 \text{ m/s}^2\) and \(h = 1.0 \text{ m}\) into the equation. |
| 10 | \[\boxed{3.7 \text{ m/s}}\] | Calculate the final velocity \(v\) at the bottom of the incline. |
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A box of mass \( 20 \) \( \text{kg} \) moves to the right on a horizontal frictionless surface with a speed of \( 4.0 \) \( \text{m/s} \). The box collides with and remains attached to one end of a spring of negligible mass whose other end is fixed to a wall. After the collision, the spring compresses a maximum distance of \( 0.50 \) \( \text{m} \), and the box then oscillates back and forth.
Ball \(A\) of mass \(m\) is dropped from a building of height \(H\). Ball \(B\) of mass \(1.7 \, \text{m}\) is dropped from a building of height \(1.7H\). Using energy, what the ratio of \(v_A\) to \(v_B\) (final velocity of ball \(A\) to final velocity of ball \(B\)). Air resistance is negligible.
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance \(d\). Ignore friction and assume that an equal force \(F\) is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
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.
A uniform, solid, \( 100 \) \( \text{kg} \) cylinder with a diameter of \( 1.0 \) \( \text{m} \) is mounted so it is free to rotate about a fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A \( 10 \) \( \text{kg} \) block is hung from a very light, thin cord wrapped around the cylinder’s circumference. When the block is released, the cord unwinds and the block accelerates downward. What is the acceleration of the block?
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 \(25 \, \text{g}\) 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?
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 \(2 \, \text{kg}\) 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?
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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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