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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."
The figure above shows a cart of mass \( M \) accelerating to the right with acceleration \( a \). A block of mass \( m \) is pressed against the cart’s front vertical surface and is held there only by friction. The coefficient of friction between the block and the cart is \( \mu \). What is the minimum acceleration \( a \) of the cart such that the block will not fall?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
A runner is moving at \( 4 \) \( \text{m/s} \). She is opposed by magic in the form of air resistance, which exerts a constant \( 20 \) \( \text{Newtons} \) force in the direction opposite her velocity. At what rate is she using energy to remain at constant velocity?
A ski tow carries people to the top of a nearby mountain. It operates on a slope of angle \( 15.7^\circ \) of length \( 260 \) \( \text{m} \). The rope moves at a speed of \( 13.0 \) \( \text{km/h} \) and provides power for \( 54 \) riders at one time, with an average mass per rider of \( 67.0 \) \( \text{kg} \).
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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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