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| Derivation or Formula | Reasoning |
|---|---|
| \[ \text{Coordinate System: } \text{x-axis: horizontally inward (toward center)}, \quad \text{y-axis: vertically upward} \] |
This coordinate system is chosen to easily resolve forces into vertical (gravity and normal force vertical component) and horizontal (providing centripetal acceleration) components. |
| \[ \text{Weight: } Mg \quad \text{(acts vertically downward)} \] |
The gravitational force \(Mg\) acts downward and is the only force due to gravity. |
| \[ \text{Normal Force: } N \quad \text{(acts perpendicular to the banked surface)} \] |
The normal force exerted by the road acts perpendicular to the surface. Since the road is banked at an angle \(\theta\), the force can be decomposed into two components. |
| \[ N\cos\theta \text{ (vertical component)} \quad \text{and} \quad N\sin\theta \text{ (horizontal component, toward the center)} \] |
The vertical component \(N\cos\theta\) balances the weight \(Mg\), and the horizontal component \(N\sin\theta\) provides the necessary centripetal force for circular motion. |
| \[ N\sin\theta = \frac{Mv_x^2}{R} \] |
This shows that the horizontal component of the normal force is equal to the centripetal force required to keep the car moving in a circular path of radius \(R\). In this diagram there is no friction, so these are the only forces. |
| Derivation or Formula | Reasoning |
|---|---|
| \[ N\cos\theta = Mg \] |
There is no vertical acceleration, so the vertical component of the normal force balances the weight of the car. |
| \[ N\sin\theta = \frac{Mv_x^2}{R} \] |
The horizontal component of the normal force provides the necessary centripetal force \(\frac{Mv_x^2}{R}\) for circular motion. |
| \[ \frac{N\sin\theta}{N\cos\theta} = \frac{\frac{Mv_x^2}{R}}{Mg}\] |
Dividing the horizontal equation by the vertical equation cancels both \(N\) and \(M\), simplifying the relationship. |
| \[ \tan\theta = \frac{v_x^2}{Rg} \] |
This simplifies to show that \(\tan\theta\) is the ratio of the centripetal term \(\frac{v_x^2}{R}\) to the gravitational acceleration \(g\). |
| \[ v_x^2 = gR\tan\theta \] |
Rearrange the equation to solve for \(v_x^2\). |
| \[ v_x = \sqrt{gR\tan\theta} \] |
Taking the square root of both sides gives the expression for the car’s speed in terms of \(g\), \(R\), and \(\theta\). |
| \[\boxed{v_x = \sqrt{gR\tan\theta}}\] | This is the final derived expression for the speed \(v_x\) of the car on the frictionless, banked curve. |
| Derivation or Formula | Reasoning |
|---|---|
| \[ \theta = 20.0^\circ, \quad R = 30.0\,\text{m}, \quad g = 9.8\,\text{m/s}^2 \] |
Substitute the given numerical values for the bank angle, radius of curvature, and gravitational acceleration. |
| \[ \tan 20.0^\circ \approx 0.364 \] |
Calculating the tangent of \(20.0^\circ\) gives approximately \(0.364\). |
| \[ v_x = \sqrt{(9.8\,\text{m/s}^2)(30.0\,\text{m})(0.364)} \] |
Substitute \(g\), \(R\), and \(\tan\theta\) into the derived speed formula \(v_x = \sqrt{gR\tan\theta}\). |
| \[ 9.8 \times 30.0 = 294\quad \text{and} \quad 294 \times 0.364 \approx 107.0 \] |
The product of \(9.8\,\text{m/s}^2 \) and \(30.0\,\text{m}\) is \(294\), and multiplying by \(0.364\) gives approximately \(107.0\). |
| \[ v_x = \sqrt{107.0} \approx 10.35\,\text{m/s} \] |
Taking the square root of \(107.0\) yields approximately \(10.35\,\text{m/s}\), which is the speed of the car. |
| \[\boxed{v_x \approx 10.35\,\text{m/s}}\] | The final numerical answer for the speed of the car on the banked curve is boxed. |
Just ask: "Help me solve this problem."
The escape speed of an object of mass \( m \) from a planet of mass \( M \) and radius \( r \) depends on the gravitational constant and
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
A 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 ° angle. The coefficient of kinetic friction of wood µk = 0.200. What magnitude of force should you apply to cause the box to slide down at a constant speed?
A truck is traveling at 35 m/s when the driver realizes the truck as no breaks. He sees a ramp off the road, inclined at 20°, and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction).
A delivery truck is traveling north. It then turns along a leftward circular curve. This the packages in the truck to slide to the RIGHT. Which of the following is true of the net force on the packages as they are sliding?
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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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