| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \(E_{\text{total}} = KE + PE\) | Total mechanical energy is the sum of kinetic energy (KE) and potential energy (PE). |
| 2 | \(KE = \frac{1}{2}mv^2\) | Kinetic energy formula where \(m\) is mass and \(v\) is velocity. |
| 3 | \(PE = mgh\) | Potential energy formula where \(g\) is the acceleration due to gravity and \(h\) is height. |
| 4 | \(E_{\text{total}}\) is constant (ignoring air resistance) | Conservation of mechanical energy principle. |
Conclusion:
The ball has the most energy at every point of its trajectory, as the total mechanical energy remains constant.
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A block is released from rest and slides down a frictionless ramp inclined at \( 30^\circ \) from the horizontal. When the block reaches the bottom, the block-Earth system has mechanical energy \( \text{E}_i \). The experiment is repeated, but now horizontal and vertical forces of magnitude \( F \) are exerted on the block while it slides, as shown above. When the block reaches the bottom, the mechanical energy of the block-Earth system.
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 car accelerates uniformly from rest to \( 29.4 \) m/s in \( 6.93 \) s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower (\( 1 \text{ horsepower} = 745.7 \text{ Watts} \)) required to accelerate the car if:
A block of mass \( 0.5 \) \( \text{kg} \) is attached to a horizontal spring with a spring constant of \( 150 \) \( \text{N/m} \). The block is released from rest at position \( x = 0.05 \) \( \text{m} \), as shown, and undergoes simple harmonic motion, reaching a maximum position of \( x = 0.1 \) \( \text{m} \). The speed of the block when it passes through position \( x = 0.09 \) \( \text{m} \) is most nearly
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 roller coaster ride at an amusement park lifts a car of mass \( 700 \, \text{kg} \) to point \( A \) at a height of \( 90 \, \text{m} \) above the lowest point on the track, as shown above. The car starts from rest at \( A \), rolls with negligible friction down the incline and follows the track around a loop of radius \( 20 \, \text{m} \). Point \( B \), the highest point on the loop, is at a height of \( 50 \, \text{m} \) above the lowest point on the track.
Find the escape speed from a planet of mass \(6.89 \times 10^{25} \, \text{kg}\) and radius \(6.2 \times 10^{6} \, \text{m}\).
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.
A \( 1.5 \; \text{kg} \) mass attached to a spring with a force constant of \( 20.0 \; \text{N/m} \) oscillates on a horizontal, frictionless track. At \( t = 0 \), the mass is released from rest at \( x = 10.0 \; \text{cm} \). (That is, the spring is stretched by \( 10.00 \; \text{cm} \).)
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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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