| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Real forces on the car: }\{\vec{W},\vec{N}\}\] | Because the road is slick and friction is negligible, there is no friction force. The only real forces are the car’s weight \(\vec{W}=M\vec{g}\) (downward) and the normal force \(\vec{N}\) (perpendicular to the road surface). |
| 2 | \[\vec{W}=Mg\,\text{(straight down)}\] | Weight always acts vertically downward with magnitude \(Mg\). |
| 3 | \[\vec{N}\perp\text{banked surface},\quad \angle(\vec{N},\text{vertical})=\theta\] | The normal force is perpendicular to the road. Since the road is banked at angle \(\theta\) above horizontal, the normal tilts by \(\theta\) from the vertical (toward the center of the circular path). |
| 4 | \[\text{Choose axes: }\hat{x}\text{ horizontal toward center (radial inward)},\;\hat{y}\text{ vertical upward}\] | This coordinate system matches the physics: centripetal acceleration is purely horizontal inward (radial), and there is no vertical acceleration because the car stays on the surface (constant height while going around). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum F_y = Ma_y = 0\] | There is no vertical acceleration for steady circular motion on the bank (the car does not accelerate upward or downward). |
| 2 | \[N\cos\theta – Mg = 0\] | Resolve \(\vec{N}\) into components using axes from part (a): vertical component is \(N\cos\theta\) upward, weight is \(Mg\) downward. |
| 3 | \[N\cos\theta = Mg\] | From \(\sum F_y=0\), the upward and downward forces balance. |
| 4 | \[\sum F_x = Ma_x = M\frac{v^2}{R}\] | Horizontal inward (radial) acceleration is centripetal with magnitude \(v^2/R\). |
| 5 | \[N\sin\theta = M\frac{v^2}{R}\] | The only horizontal inward force is the inward component of the normal force, \(N\sin\theta\). |
| 6 | \[\frac{N\sin\theta}{N\cos\theta} = \frac{M\frac{v^2}{R}}{Mg}\] | Divide the horizontal equation by the vertical equation to eliminate \(N\) and \(M\). |
| 7 | \[\tan\theta = \frac{v^2}{Rg}\] | Simplify: \(\sin\theta/\cos\theta=\tan\theta\) and \(\left(Mv^2/R\right)/(Mg)=v^2/(Rg)\). |
| 8 | \[v^2 = Rg\tan\theta\] | Solve algebraically for \(v^2\). |
| 9 | \[\boxed{v = \sqrt{Rg\tan\theta}}\] | This is the required speed for a frictionless banked curve; mass \(M\) cancels, so the speed does not depend on \(M\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v = \sqrt{Rg\tan\theta}\] | Use the result from part (b). |
| 2 | \[v = \sqrt{(30.0\,\text{m})(9.80\,\text{m/s}^2)\tan(20.0^\circ)}\] | Substitute \(R=30.0\,\text{m}\), \(g=9.80\,\text{m/s}^2\), \(\theta=20.0^\circ\). (\(M\) is not needed because it cancels.) |
| 3 | \[v = \sqrt{(294\,\text{m}^2/\text{s}^2)\tan(20.0^\circ)}\] | Compute \((30.0)(9.80)=294\). |
| 4 | \[v \approx \sqrt{(294)(0.364)}\] | Use \(\tan(20.0^\circ)\approx 0.364\). |
| 5 | \[v \approx \sqrt{107}\] | Compute \((294)(0.364)\approx 107\). |
| 6 | \[\boxed{v \approx 10.3\,\text{m/s}}\] | Take the square root: \(\sqrt{107}\approx 10.3\), giving the car’s speed on the banked, frictionless curve. |
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A roller coaster ride at an amusement park lifts a car of mass \( 700 \, \text{kg} \) to point \( A \) at a height of \( 90 \, \text{m} \) above the lowest point on the track, as shown above. The car starts from rest at \( A \), rolls with negligible friction down the incline and follows the track around a loop of radius \( 20 \, \text{m} \). Point \( B \), the highest point on the loop, is at a height of \( 50 \, \text{m} \) above the lowest point on the track.
Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of the ball relative to the ground. Describe its motion relative to the wagon.
If the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
You are pushing a heavy box across a rough floor. When you are initially pushing the box and it is accelerating,
Which pulls harder gravitationally, the Earth on the Moon, or the Moon on the Earth? Which accelerates more?
A simple Atwood’s machine remains motionless when equal masses \(M\) are placed on each end of the chord. When a small mass \(m\) is added to one side, the masses have an acceleration \(a\). What is \(M\)? You may neglect friction and the mass of the cord and pulley.
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 block hangs from the ceiling by a massless rope. A \( 3.0 \, \text{kg} \) block is attached to the first block and hangs below it on another piece of massless rope. The tension in the top rope is \( 63.0 \, \text{N} \).
If an object is moving, is it possible for the net force acting on it to be zero? Explain.
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
\(v = \sqrt{Rg\tan\theta}\)
\(10.3\,\text{m/s}\)
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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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