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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]E_{\text{total}} = PE_{\text{max}} = KE_{\text{max}}[/katex] | Total mechanical energy is constant, equal to maximum PE (at highest point) and maximum KE (at lowest point). |
| 2 | [katex]E_{\text{total}} = KE + PE[/katex] | At any point, total mechanical energy is the sum of KE and PE. |
| 3 | [katex]PE = mgh[/katex] | Potential energy at any point, where [katex]h[/katex] is height above the lowest point. |
| 4 | [katex]mgh = \frac{1}{2} E_{\text{total}}[/katex] | When KE equals PE, each is half the total mechanical energy. |
| 5 | [katex]h = L(1 – \cos(\theta))[/katex] | Height in terms of the pendulum length [katex]L[/katex] and angle [katex]\theta[/katex]. |
| 6 | [katex]L(1 – \cos(\theta)) = \frac{1}{2} L[/katex] | Substitute [katex]h[/katex] into the equation from Step 4, using [katex]E_{\text{total}} = mgL[/katex] (max PE). |
| 7 | [katex]1 – \cos(\theta) = \frac{1}{2}[/katex] | Simplify the equation. |
| 8 | [katex]\cos(\theta) = \frac{1}{2}[/katex] | Rearrange the equation to solve for [katex]\cos(\theta)[/katex]. |
| 9 | [katex]\theta \approx 60^\circ[/katex] | Solve for [katex]\theta[/katex], knowing that [katex]\cos^{-1}\left(\frac{1}{2}\right) \approx 60^\circ[/katex]. |
Therefore, the angle [katex]\theta[/katex] at which the kinetic energy equals the potential energy is approximately [katex]\boxed{60^\circ}[/katex] from the VERTICAL. From the horizontal this angle would be 90°-60° = 30° which is less than 45°. Note the questions ask for the angle from the horizontal.
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In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?
How does the speed \(v_1\) of a block \(m\) reaching the bottom of slide 1 compare with \(v_2\), the speed of a block \(2m\) reaching the end of slide 2? The blocks are released from the same height.
A block is attached to a horizontal spring and is initially at rest at the equilibrium position \( x = 0 \), as shown in Figure \( 1 \). The block is then moved to position \( x = -A \), as shown in Figure \( 2 \), and released from rest, undergoing simple harmonic motion. At the instant the block reaches position \( x = +A \), another identical block is dropped onto and sticks to the block, as shown in Figure \( 3 \). The two–block–spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy \( E_{\text{tot},2} \) of the two–block–spring system after the collision to the total mechanical energy \( E_{\text{tot},1} \) of the one–block–spring system before the collision?
In \(3.0 \, \text{minutes}\), a ski lift raises \(10\) skiers at constant speed to a height of \(85 \, \text{m}\). The ski lift is \(55^\circ\) above the horizontal and the average mass of each skier is \(67.5 \, \text{kg}\). What is the average power provided by the tension in the cable pulling the lift?
A mechanic pushes a [katex]2500 \, \text{kg}[/katex] car from rest to a final speed [katex]v[/katex] by doing [katex]5.0 \times 10^3 \, \text{J}[/katex] of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
A \(0.5 \, \text{kg}\) cart, on a frictionless \(2 \, \text{m}\) long table, is being pulled by a \(0.1 \, \text{kg}\) mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls \(0.5 \, \text{m}\), calculate the speed of the cart on the table. Use ONLY forces and energy.
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
A big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s2. How far up does it move?
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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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