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UBQ Credits
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}} [/katex] | Total mechanical energy is the sum of kinetic and potential energy. |
| 2 | [katex] E_{\text{kinetic}} = \frac{1}{2}mv^2 [/katex] | Kinetic energy formula, where [katex] E_{\text{kinetic}} [/katex] is kinetic energy, [katex] m [/katex] is mass, and [katex] v [/katex] is velocity. |
| 3 | [katex] E_{\text{potential}} = mgh [/katex] | Potential energy formula, where [katex] E_{\text{potential}} [/katex] is potential energy, [katex] h [/katex] is height, and [katex] g [/katex] is gravitational acceleration. |
| 4 | At the top, [katex] E_{\text{total}} = E_{\text{potential}} [/katex] | Initially, all energy is potential energy since velocity is zero. |
| 5 | At the bottom, [katex] E_{\text{total}} = E_{\text{kinetic}} [/katex] | At the bottom, all energy is converted to kinetic energy, assuming negligible air resistance. |
| 6 | [katex] mgh = \frac{1}{2}mv^2 [/katex] | Equating potential energy at the top with kinetic energy at the bottom. |
| 7 | [katex] 2gh = v^2 [/katex] | Cancel [katex] m [/katex] and rearrange the equation. |
| 8 | [katex] v = \sqrt{2gh} [/katex] | Take the square root to find [katex] v [/katex]. |
| 9 | [katex] v_A = \sqrt{2gH} [/katex] | Apply the formula to ball A, dropped from height [katex] H [/katex]. |
| 10 | [katex] v_B = \sqrt{2g \cdot 3.5H} [/katex] | Apply the formula to ball B, dropped from height [katex] 3.5H [/katex]. |
| 11 | [katex] \frac{v_A}{v_B} = \frac{\sqrt{2gH}}{\sqrt{7gH}} [/katex] | Compare the velocities of the two balls. |
| 12 | [katex] \boxed{\frac{v_A}{v_B} = \sqrt{\frac{2}{7}}} [/katex] | Simplify to find the ratio. |
The derivation uses energy principles to arrive at the final velocity formula, and the ratio of velocities of ball A to ball B is [katex] \sqrt{\frac{2}{7}} [/katex].
Just ask: "Help me solve this problem."
A bullet moving with an initial speed of [katex] v_o [/katex] strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height [katex] h [/katex]. Which of the following statements is true of the collision.
A skier with a mass of 58 kg glides up a snowy incline that forms an angle of 28 degrees with the horizontal. The skier initially moves at a speed of 7.2 m/s. After traveling a distance of 2.3 meters up the slope, the skier’s speed reduces to 3.8 m/s.
In 3.0 minutes, a ski lift raises 10 skiers at constant speed to a height of 85 m. The ski lift is 55° above the horizontal and the average mass of each skier is 67.5 kg. What is the average power provided by the tension in the cable pulling the lift?
It takes 4 seconds for an individual to push a 70 kg box up a 5m long, 12° ramp. The box starts from rest and achieves a speed of 2.5 m/s at the top. Friction does 350 J of work during its ascent. Calculate the power output of the individual pushing the box.
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
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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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