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| 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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Three students are pulling on a bag of skittles. Each is pulling with a horizontal force. If student 1 pulls Eastward with [katex]170 \, \text{N}[/katex], student 2 pulls Southward with [katex]100 \, \text{N}[/katex] and student 3 pulls with [katex]200 \, \text{N}[/katex] at an angle of [katex]20^\circ [/katex] west of north, what is the net force caused by the three students on the bag of skittles?
A body starting from rest moves along a straight line under the action of a constant force. After traveling a distance \( d \) the speed of the body is \( v \). The speed of the body when it has travelled a distance \( \dfrac{d}{2} \) from its initial position is
You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, \(12 \, \text{m}\) above, in a time of \(6 \, \text{s}\). The scale reads \(800 \, \text{N}\). What is the mass of the person?
Imagine a hypothetical planet that has two moons. Moon \(\#1\) is in a circular orbit of radius \(R\) and has a mass \(M\).
The escape speed of an object of mass \( m \) from a planet of mass \( M \) and radius \( r \) depends on the gravitational constant and
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?
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 0.33 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
Three blocks of masses \( 1.0 \, \text{kg} \), \( 2.0 \, \text{kg} \), and \( 4.0 \, \text{kg} \) are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following.
What force would have to be applied to start a \(12.3 \, \text{kg}\) wood block moving on a surface with a static coefficient of friction of \(0.438\)?
There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
\(v = \sqrt{Rg\tan\theta}\)
\(10.3\,\text{m/s}\)
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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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