| 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 \(10 \, \text{kg}\) box is pushed to the right by an unknown force at an angle of \(25^\circ\) below the horizontal while a friction force of \(50 \, \text{N}\) acts on the box as well. The box accelerates from rest and travels a distance of \(4 \, \text{m}\) where it is moving at \(3 \, \text{m/s}\).
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of \( 38.0^\circ \) with respect to the horizontal. Suppose a wooden crate with a mass of \( 25.0 \) \( \text{kg} \) is placed on Baldwin Street. An additional force of \( 59 \) \( \text{N} \) must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is \( 0.599 \), what is the magnitude of the frictional force?
A \(1509 \, \text{g}\) wood block is being pulled by the force meter at a constant velocity. Using the graph above, find:
A box having a mass of \( 1.5 \) \( \text{kg} \) is accelerated across a table at \( 1.5 \) \( \text{m/s}^2 \). The coefficient of kinetic friction on the box is \( 0.3 \).
A \( 25.0 \) \( \text{kg} \) block is initially at rest on a horizontal surface. A horizontal force of \( 75.0 \) \( \text{N} \) is required to set the block in motion, after which a horizontal force of \( 60.0 \) \( \text{N} \) is required to keep the block moving with constant speed.
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
A force \(F\) is used to hold a block of mass \(m\) on an incline as shown in the diagram above. The plane makes an angle of \(\theta\) with the horizontal and \(F\) is perpendicular to the plane. The coefficient of friction between the plane and the block is \(\mu\). What is the minimum force \(F\) necessary to keep the block at rest?
A block of mass \( 4.0 \) \( \text{kg} \) rests on an inclined plane. The coefficient of static friction between the block and the plane \( \mu_s \) is \( 0.4 \). Which of the following gives the angle of inclination at which the block will start to slide?
A block rests on a flat plane inclined at an angle of \(30^\circ\) with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
Why is the stopping distance of a truck much shorter than for a train going the same speed? Hint: try deriving a formula or stopping distance.
\( 5.33 \, \text{m/s}^2 \). 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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