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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]h = s \sin(\theta)[/katex] | To find the vertical height [katex]h[/katex] gained by the skier on the slope (where [katex]s = 695[/katex] meters, [katex]\theta = 34^\circ[/katex]), we use the sine component of the inclined angle because height is opposed to gravity and directly vertical. |
| 2 | [katex]h = 695 \sin(34^\circ)[/katex] | Calculate [katex]h[/katex] by substituting the values of [katex]s[/katex] and [katex]\theta[/katex]. The sine of [katex]34^\circ[/katex] (from calculator or trigonometric table) is used to obtain [katex]h[/katex]. |
| 3 | [katex]PE = mgh[/katex] | Calculate the potential energy ([katex]PE[/katex]) needed to lift a skier to height [katex]h[/katex]. Here, [katex]m[/katex] is mass (72 kg), [katex]g[/katex] is the acceleration due to gravity (approximately [katex]9.81 \, \text{m/s}^2[/katex]), and [katex]h[/katex] is the height obtained in the previous step. |
| 4 | [katex]PE = 72 \cdot 9.81 \cdot 695 \sin(34^\circ)[/katex] | Substitute the values into the potential energy formula to find the energy required to transport one skier to the top of the slope. |
| 5 | [katex]P = \frac{PE \cdot R}{t}[/katex] | Calculate the power ([katex]P[/katex]) needed to transport [katex]R[/katex] riders per minute. [katex]t[/katex] is the time in seconds; for one minute, [katex]t = 60[/katex] seconds. |
| 6 | [katex]P = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60}[/katex] | Substitute the values to find the power required for ferrying 5 riders per minute. |
| 7 | [katex]P_{\text{total}} = \frac{P}{\text{efficiency}}[/katex] | To find the total average power ([katex]P_{\text{total}}[/katex]) supplied by the motor, we adjust for the efficiency of the ski lift, which only uses 65% of the energy supplied to overcome work against friction. |
| 8 | [katex]P_{\text{total}} = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60 \cdot 0.65}[/katex] | Calculate [katex]P_{\text{total}}[/katex] by considering the efficiency. This power value indicates the power that must be supplied by the motor. |
| 9 | [katex]P_{\text{total}} \approx 35,192.76 \, \text{Watt}[/katex] | Complete the calculations to get the final answer in Watts (rounded to sensible precision). |
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A 2 kg model rocket is launched with a thrust force of 275 N and reaches a height of 90 m, moving at 150 m/s at its peak. What is the average air resistance force acting on the rocket during its ascent?
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
A particle of mass m slides down a fixed, frictionless sphere of radius R, starting from rest at the top.
In terms of m, g, R, and O, determine each of the following for the particle while it is sliding on the sphere.
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle [katex] \theta [/katex] with the vertical. The change in potential energy of the bob during the displacement is:
35,192.76 Watts
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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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