| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a_{\text{level}} = \mu g \quad \Longrightarrow \quad \mu = \frac{3.80}{g}\] | On a level road the maximum deceleration is provided entirely by static friction. Since the car decelerates at \(-3.80 \, \text{m/s}^2\) (in magnitude \(3.80\, \text{m/s}^2\)), we can write \(a_{\text{level}} = \mu g\) and solve for the friction coefficient \(\mu\). |
| 2 | \[N = mg\cos(9.3^\circ)\] | On an incline the normal force is reduced to \(mg\cos(9.3^\circ)\); this reduction affects the maximum static friction available. |
| 3 | \[f_{\text{max}} = \mu mg\cos(9.3^\circ)\] | The maximum frictional force that can be provided on the incline is given by \(\mu N\), which is \(\mu mg\cos(9.3^\circ)\). |
| 4 | \[a_{\text{friction}} = \mu g\cos(9.3^\circ)\] | Dividing the maximum frictional force by the mass gives the maximum deceleration contribution from friction on the inclined plane. |
| 5 | \[a_{\text{gravity}} = g\sin(9.3^\circ)\] | When the car is moving uphill, the gravitational component along the slope, \(g\sin(9.3^\circ)\), also works to decelerate the car (acting downhill). |
| 6 | \[a_{\text{uphill}} = \mu g\cos(9.3^\circ) + g\sin(9.3^\circ)\] | The net deceleration is the sum of the deceleration from friction and the deceleration due to the gravitational component along the incline. |
| 7 | \[a_{\text{uphill}} = \left(\frac{3.80}{g}\right)g\cos(9.3^\circ) + g\sin(9.3^\circ) = 3.80\cos(9.3^\circ) + g\sin(9.3^\circ)\] | Substitute \(\mu = \frac{3.80}{g}\) from Step 1 into the net deceleration formula. |
| 8 | \[a_{\text{uphill}} = 3.80\cos(9.3^\circ) + 9.8\sin(9.3^\circ)\] | Using \(g = 9.8\, \text{m/s}^2\), the expression now contains numerical values and the trigonometric functions of \(9.3^\circ\). |
| 9 | \[a_{\text{uphill}} \approx 3.80(0.987) + 9.8(0.161) \approx 3.75 + 1.58 \approx 5.33 \, \text{m/s}^2\] | Evaluating \(\cos(9.3^\circ) \approx 0.987\) and \(\sin(9.3^\circ) \approx 0.161\) gives a net deceleration of about \(5.33\, \text{m/s}^2\). Since the deceleration is opposite to the direction of motion, it is expressed as a negative acceleration. |
| 10 | \[\boxed{a_{\text{uphill}} = -5.33 \, \text{m/s}^2}\] | This is the final result: when moving uphill on a \(9.3^\circ\) incline with the same static friction coefficient, the car decelerates at approximately \(-5.33 \, \text{m/s}^2\). |
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A \(25.0 \, \text{kg}\) box is released on a \(23.5^\circ\) incline and accelerates down the incline at \(0.35 \, \text{m/s}^2\). Find the friction force impeding its motion. What is the coefficient of kinetic friction?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
A \( 1 \) \( \text{kg} \) mass on a \( 37^{\circ} \) incline is connected to a \( 3.0 \) \( \text{kg} \) mass on a horizontal surface, as shown. The surfaces and the pulley are frictionless. If \( F = 12 \) \( \text{N} \):
A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
A child on a sled reaches the bottom of a hill with a velocity of \( 10.0 \, \text{m/s} \) and travels \( 25.0 \, \text{m} \) along a horizontal straightaway to a stop. If the child and sled together have a mass of \( 60.0 \, \text{kg} \), what is the average retarding force on the sled on the horizontal straightaway?
A block is given a brief push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
A \(6 \, \text{kg}\) cube rests against a compressed spring with a force constant of \(1{,}800 \, \text{N/m}\), initially compressed by \(0.3 \, \text{m}\). Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of \(\mu_k = 0.12\) for \(3 \, \text{m}\), then ascends a \(12^\circ\) slope, stopping after \(4.5 \, \text{m}\). Determine the coefficient of kinetic friction on the slope.
A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is \(10.0^\circ\). What is the coefficient of kinetic friction between the sled and the hill’s surface?
\( 5.33 \, \text{m/s}^2 \). A negative number indicating deceleration is acceptable.
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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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