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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."
A 10kg box is pushed to the right by an unknown force at an angle of 25° below the horizontal while a friction force of 50 N acts on the box as well. The box accelerates from rest and travels a distance of 4 m where it is moving at 3 m/s. Solve the following without the use of energy.
The Moon does not crash into the Earth because:
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
A ball falls straight down through the air under the influence of gravity. There is a retarding force \(F\) on the ball with magnitude given by \(F=bv\), where \(v\) is the speed of the ball and \(b\) is a positive constant. The ball reaches a terminal velocity after a time \(t\). The magnitude of the acceleration at time \(t/2\) is
The cart with mass \( M = 3 \, \text{kg} \) is pulled by a massless string and moving on a horizontal track. A weight with mass \( m = 1 \, \text{kg} \) is hung from the other end of the string through a pulley system. Due to the gravitational force acting on the weight of mass \( m \), the cart is accelerated to the left. Find the tension in the string.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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