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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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Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance d. Ignore friction and assume that an equal force F is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
A bullet at speed [katex] v_0 [/katex] trikes 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 h. Which of the following statements is true?
The box in the diagram is sliding to the right across a horizontal table, under the influence of the forces shown. Which force(s) is doing negative work on the box?
A ball is thrown straight up. At what point does the ball have the most energy?
A proton (mp = 1.67 x10-27 kg) is being accelerated along a straight line at 3.6 ×1015 m/s2 in a machine. The proton has an initial speed of 2.4 x107 m/s and travels 3.5 cm.
35,192.76 Watts
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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