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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{parallel}} = mg\sin(\theta)[/katex] | Force parallel to the incline, where [katex]m[/katex] is mass, [katex]g[/katex] is acceleration due to gravity, and [katex]\theta[/katex] is the incline angle. |
| 2 | [katex]F_{\text{normal}} = mg\cos(\theta)[/katex] | Normal force, perpendicular to the incline. |
| 3 | [katex]F_{\text{friction}} = \mu_k F_{\text{normal}}[/katex] | Frictional force, where [katex]\mu_k[/katex] is the coefficient of kinetic friction. |
| 4 | [katex]F_{\text{friction}} = 0.250 \times mg\cos(25^\circ)[/katex] | Substitute values for [katex]\mu_k[/katex] and [katex]F_{\text{normal}}[/katex]. |
| 5 | [katex]F_{\text{pull}} = F_{\text{parallel}} + F_{\text{friction}}[/katex] | Total force to pull the crate includes parallel force and frictional force. |
| 6 | [katex]F_{\text{pull}} = mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ)[/katex] | Combine parallel and frictional forces. |
| 7 | [katex]W_{\text{output}} = F_{\text{parallel}} \times d[/katex] | Work output, where [katex]F_{\text{parallel}}[/katex] is the force parallel to the incline and [katex]d[/katex] is the distance. |
| 8 | [katex]W_{\text{output}} = mg\sin(25^\circ) \times 2.5[/katex] | Work done against gravity. |
| 9 | [katex]W_{\text{input}} = F_{\text{pull}} \times d[/katex] | Work input, force to pull the crate. |
| 10 | [katex]W_{\text{input}} = \left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5[/katex] | Work done to pull the crate including overcoming friction. |
| 11 | [katex]\text{Efficiency} = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100%[/katex] | Efficiency formula. |
| 12 | [katex]\text{Efficiency} = \frac{mg\sin(25^\circ) \times 2.5}{\left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5} \times 100%[/katex] | Substitute [katex]W_{\text{output}}[/katex] and [katex]W_{\text{input}}[/katex]. |
| 13 | [katex]\text{Efficiency} = \frac{\sin(25^\circ)}{\sin(25^\circ) + 0.250 \times \cos(25^\circ)} \times 100%[/katex] | Simplify by canceling [katex]mg[/katex] and [katex]d[/katex]. |
| 14 | [katex]\text{Efficiency} = 65.10\%[/katex] | Calculated efficiency. |
Just ask: "Help me solve this problem."
A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at [katex] 470 \, \frac{rad}{s^2}[/katex]. The ball is released after [katex]\frac{3}{4} [/katex] of a revolution. How high does it go above the center of the wheel?
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
An average adult elephant \( (5000 \, \text{kg}) \) is strapped to a spring, which is then pulled \( 2 \, \text{meters} \) away from its equilibrium position and released. The elephant starts oscillating back and forth with a period of \( 10 \) seconds.
A spring launches a 4 kg block across a frictionless horizontal surface. The block then ascends a 30° incline with a kinetic friction coefficient of 0.25, stopping after 55 m on the incline. If the spring constant is 800 N/m, find the initial compression of the spring. Disregard friction while in contact with the spring.
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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] |
General Metric Conversion Chart
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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