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| 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. |
Just ask: "Help me solve this problem."
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
A box is sliding down an incline at a constant speed of \( 2 \) \( \text{m s}^{-1} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \) \( \text{N} \). What is the force of gravity acting on the box?
What force would have to be applied to start a 12.3 kg wood block moving on a surface with a static coefficient of friction of 0.438?
A car of mass \( M \) moves around a circularly banked curve on a freeway off-ramp. The off-ramp has a radius of curvature \( R \) and is raised to an angle \( \theta \) from the horizontal. The road is slick, and friction is negligible.
When the brakes of an automobile are applied, the road exerts the greatest retarding force
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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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