| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \(E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}}\) | Total mechanical energy in a spring-mass system is conserved, comprising kinetic and potential energy. |
| 2 | \(E_{\text{potential}} = \frac{1}{2}kx^2\) | Potential energy in a spring, where \(k\) is the spring constant and \(x\) is the displacement from equilibrium. |
| 3 | \(E_{\text{kinetic}} = \frac{1}{2}mv^2\) | Kinetic energy of the mass, where \(m\) is the mass and \(v\) is the velocity. |
| 4 | \(E_{\text{total, initial}} = \frac{1}{2}kx^2\) | Total energy initially (at maximum stretch) is all potential energy. Given: \(k = 25 , \text{N/m}, x = 52.0 , \text{cm} = 0.52 , \text{m}\). |
| 5 | \(E_{\text{total, halfway}} = E_{\text{kinetic, halfway}} + E_{\text{potential, halfway}}\) | Total energy halfway to equilibrium. |
| 6 | \(E_{\text{potential, halfway}} = \frac{1}{2}k\left(\frac{x}{2}\right)^2\) | Potential energy halfway to equilibrium (\(x/2\)). |
| 7 | \(E_{\text{total, initial}} = E_{\text{total, halfway}}\) | Conservation of mechanical energy. |
| 8 | \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \frac{1}{2}k\left(\frac{x}{2}\right)^2\) | Equate initial and halfway total energies. |
| 9 | \(v = \sqrt{\frac{kx^2 – k\left(\frac{x}{2}\right)^2}{m}}\) | Solve for \(v\). Given: \(m = 80 , \text{g} = 0.080 , \text{kg}\). |
Let’s calculate the velocity of the mass when it is halfway to the equilibrium position.
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 10 | \(v = 7.96 , \text{m/s}\) | Velocity of the mass halfway to the equilibrium position. |
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A \(90 \, \text{kg}\) individual is cycling up a hill inclined at \(30^\circ\) on a \(12 \, \text{kg}\) bicycle. The hill is quite steep, and the coefficient of static friction is \(0.85\). The cyclist ascends \(12 \, \text{m}\) up the hill and then pauses at the summit. They then start descending from rest and travel \(9 \, \text{m}\) before firmly applying the brakes, causing the wheels to lock.
A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of \(0.80 \, \text{m}\) above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is \(0.35 \, \text{m}\). The mass of the marble is \(0.050 \, \text{kg}\). Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.
You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of \(1.25 \, \text{m}\) in a time of \(0.84 \, \text{s}\). The readout on the display indicates that the average power you are producing is \(76 \, \text{W}\). What is the magnitude of the force that you exert on the handle?
A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
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?

An object of mass \( m = 3.0 \) \( \text{kg} \) is attached to one end of a string with negligible mass and length \( L = 0.80 \) \( \text{m} \). The object is released from rest at time \( t = 0 \), when the string is horizontal. At time \( t = t_1 \) the object is at the location shown in the figure, where the string is vertical. Which of the following is most nearly the magnitude of the tension in the string at time \( t = t_1 \)?
A skier with a mass of \(58 \, \text{kg}\) glides up a snowy incline that forms an angle of \(28^\circ\) with the horizontal. The skier initially moves at a speed of \(7.2 \, \text{m/s}\). After traveling a distance of \(2.3 \, \text{m}\) up the slope, the skier’s speed reduces to \(3.8 \, \text{m/s}\).
When can the motion of a pendulum be modeled as simple harmonic motion?
7.96 m/s
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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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