| 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. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
A race car travels in a circular track of radius \( 200 \) \( \text{m} \). If the car moves with a constant speed of \( 80 \) \( \text{m/s} \),
A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).
A car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?
A communications satellite orbits the Earth at an altitude of \(35{,}000 \, \text{km}\) above the Earth’s surface. Take the mass of Earth to be \(6 \times 10^{24} \, \text{kg}\) and the radius of Earth to be \(6.4 \times 10^6 \, \text{m}\). What is the satellite’s velocity?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
According to Newton’s third law, each team in a tug of war pulls with equal force on the other team. What, then, determines which team will win?
A \(1.5 \, \text{kg}\) object is located at a distance of \(1.7 \times 10^{6} \, \text{m}\) from the center of a larger object whose mass is \(7.4 \times 10^{22} \, \text{kg}\).
A coffee cup on the dashboard of a car slides forward when the driver decelerates from \(45 ~ \frac{\text{km}}{\text{hr}}\) to rest in \(3.5 \, \text{s}\) or less. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are completely horizontal.
For linear motion the term “inertia” refers to the same physical concept of
A 2.0 kg ball on the end of a 0.65 m long string is moving in a vertical circle. At the bottom of the circle, its speed is 4.0 m/s. Find the tension in the string.
\(v = \sqrt{Rg\tan\theta}\)
\(10.3\,\text{m/s}\)
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
