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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}} – E_{\text{friction}} [/katex] | Total mechanical energy considering kinetic energy, potential energy, and energy lost to friction. |
| 2 | [katex] E_{\text{kinetic}} = \frac{1}{2}mv^2 [/katex] | Kinetic energy formula. |
| 3 | [katex] E_{\text{potential}} = mgh [/katex] | Potential energy formula, where [katex] h [/katex] is height. |
| 4 | [katex] E_{\text{friction}} = f_k \cdot d [/katex] | Energy lost to friction, where [katex] f_k [/katex] is kinetic friction force, and [katex] d [/katex] is distance. |
| 5 | [katex] f_k = \mu_k N [/katex] | Kinetic friction force formula, where [katex] \mu_k [/katex] is coefficient of kinetic friction, and [katex] N [/katex] is normal force. |
| 6 | [katex] N = mg\cos(\theta) [/katex] | Normal force on an incline, where [katex] \theta [/katex] is the angle of the incline. |
| 7 | [katex] E_{\text{friction}} = \mu_k mg\cos(\theta) \cdot d [/katex] | Combine steps 4 and 5. |
| 8 | [katex] mgh = \frac{1}{2}mv^2 + \mu_k mg\cos(\theta) \cdot d [/katex] | Equate initial potential energy to kinetic energy and energy lost to friction. |
| 9 | [katex] v = \sqrt{\frac{2(mgh – \mu_k mg\cos(\theta) \cdot d)}{m}} [/katex] | Solve for [katex] v [/katex]. |
For both masses, the variables [katex] h, \mu_k, g, \theta, [/katex] and [katex] d [/katex] are constant. The term [katex] mg [/katex] in the friction energy cancels out when solving for [katex] v [/katex]. Thus, the final speed [katex] v [/katex] does not depend on the mass.
Just ask: "Help me solve this problem."
A 6 kg cube rests against a compressed spring with a force constant of 1,800 N/m, initially compressed by 0.3 m. Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of 0.12 for 3 m, then ascends a 12° slope, stopping after 4.5 m. Determine the coefficient of kinetic friction on the slope.
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
A horizontal force of 110 N is applied to a 12 kg object, moving it 6 m on a horizontal surface where the kinetic friction coefficient is 0.25. The object then slides up a 17° inclined plane. Assuming the 110 N force is no longer acting on the incline, and the coefficient of kinetic friction there is 0.45, calculate the distance the object will slide on the incline.
A ski lift carries skiers along a 695 meter slope inclined at 34°. To lift a single rider, it is necessary to move 72 kg of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If 65 percent of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
A 0.5 kg cart, on a frictionless 2 m long table, is being pulled by a 0.1 kg mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls 0.5 m, calculate the speed of the cart on the table. Use ONLY forces and energy.
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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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