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| 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 \(-5.33 \, \text{m/s}^2\). |
Just ask: "Help me solve this problem."
A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
A person is running on a track. Which of the following forces propels the runner forward?
What would your bathroom scale read if you weighed yourself on an inclined plane? Assume the mechanism functions properly, even at an angle.
A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a \( 30^\circ \) rough incline (\( \mu_k = 0.4 \)) of distance \( d \), with an initial speed of \( 5 \) \( \text{m/s} \), and then it slides back down.
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] |
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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