| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]KE_{\text{initial}} = \frac{1}{2} m v^2[/katex] | Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here [katex]v = 12.3 \text{ m/s}[/katex]. |
| 2 | [katex]W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d[/katex] | The work done by friction, where [katex] \mu_k = 0.650[/katex] is the coefficient of kinetic friction, [katex] g = 9.8 \text{ m/s}^2[/katex] is acceleration due to gravity, [katex] \theta = 18^\circ[/katex], and [katex]d[/katex] is the distance the vehicle slides. |
| 3 | [katex]W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d[/katex] | The work done by gravity while the vehicle moves down the incline. |
| 4 | [katex]KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}}[/katex] | By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity. |
| 5 | [katex]\frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d[/katex] | Substitute expressions from steps 1, 2, and 3 into the work-energy equation. |
| 6 | [katex]d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)}[/katex] | Solve for [katex]d[/katex], distance the vehicle slides. Notice that mass [katex]m[/katex] cancels out. |
| 7 | [katex]d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)}[/katex] | Substitute numerical values for [katex]v[/katex], [katex]\mu_k[/katex], [katex]g[/katex], and [katex]\theta[/katex] to find the value of [katex]d[/katex] that represents the distance the vehicle slides until it stops. |
| 8 | [katex]d = 8.32 \,\text{m}[/katex] | Calculated value. |
To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle [katex] \theta [/katex] the shorter the distance traveled.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]d’ = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)}[/katex] | Recalculate the distance with the increased angle of [katex]1.5 \times 18^\circ = 27^\circ[/katex]. |
| 2 | [katex]d’ = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)}[/katex] | Using the previous formula of [katex]d[/katex], we express the new sliding distance [katex]d'[/katex] in terms of the old distance [katex]d[/katex]. |
| 3 | [katex]d’ = 7.41 \, \text{meters}[/katex] | At an angle of [katex]27^\circ[/katex] the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters. |
| 3 | [katex]\frac{d’}{d}[/katex] | The ratio [katex]\frac{d’}{d}[/katex] shows how much further the vehicle would slide relative to [katex]d[/katex]. |
| 4 | [katex]\frac{7.41}{8.32} = .89[/katex] | Thus the new distance [katex]d’ = .89d[/katex] |
Part (c):
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d[/katex] | The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion. |
| 2 | [katex]\frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d[/katex] | Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force. |
| 3 | [katex]d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)}[/katex] | Solve for the distance the vehicle would slide up the incline. |
| 4 | [katex]d_{\text{up}} < d[/katex] | The distance [katex]d_{\text{up}}[/katex] will be lesser than [katex]d[/katex] since gravity now acts against the motion, reducing the sliding distance relative to sliding down. |
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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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