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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]h = L – L \cos(\theta)[/katex] | Calculate the vertical height the pendulum rises. This uses the initial angle of 60 degrees and the length of the pendulum. |
| 2 | [katex]PE_{\text{top}} = mgh[/katex] | Calculate the potential energy at the highest point. Potential energy is defined as the product of mass, gravitational acceleration, and height. |
| 3 | [katex]KE_{\text{bottom}} = \frac{1}{2}mv^2[/katex] | Calculate the kinetic energy at the bottom. Kinetic energy is defined as half the product of mass and the square of the velocity. |
| 4 | [katex]PE_{\text{top}} = KE_{\text{bottom}}[/katex] | Apply the conservation of mechanical energy. The potential energy at the highest point is converted to kinetic energy at the lowest point. |
| 5 | [katex]mgh = \frac{1}{2}mv^2[/katex] | Set the potential energy equal to the kinetic energy to find the relationship between the height and the speed at the lowest point. |
| 6 | [katex]v = \sqrt{2gh}[/katex] | Solve for the velocity. Here, mass cancels out, showing that the velocity does not depend on the mass of the pendulum. |
| 7 | [katex]v = \sqrt{2g(L – L\cos(\theta))}[/katex] | Substitute the height equation into the velocity equation. This gives the velocity in terms of the pendulum length and the angle. |
| 8 | [katex]v = \sqrt{2gL(1 – \cos(\theta))}[/katex] | Final simplification to find the expression for the velocity at the lowest point of the swing. |
Final substitution
θ₀ = 60°, cos 60° = ½
v = √[2gL(1 − ½)]
= √(2gL · ½)
= √(gL)
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Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star?
A runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.
A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
A lighter car and a heavier truck, each traveling to the right with the same speed [katex] v [/katex] hit their brakes. The retarding frictional force F on both cars turns out to be constant and the same. After both vehicles travel a distance [katex] D [/katex] (and both are still moving), which of the following statements is true?
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
A man weighing \( 700 \) \( \text{N} \) and a woman weighing \( 400 \) \( \text{N} \) have the same momentum. What is the ratio of the man’s kinetic energy \( K_m \) to that of the woman \( K_w \)?
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?

A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
A block attached to spring demonstrates simple harmonic motion about its equilibrium position with amplitude [katex] A [/katex] and angular frequency [katex] \omega [/katex]. What is the maximum magnitude of the block’s velocity?
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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