| 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 person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
An \(80 \, \text{kg}\) person sits on a swing ride that moves in a horizontal circle. The swing is suspended by a chain of length \(8 \, \text{m}\). While in motion, the chain makes a \(40^\circ\) angle with the horizontal. What is the speed of the person?
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
Two balls have their centers \( 2.0 \) \( \text{m} \) apart. One ball has a mass of \( 8.0 \) \( \text{kg} \). The other has a mass of \( 6.0 \) \( \text{kg} \). What is the gravitational force between them?

A neighbor’s child wants to go to a carnival to experience the wild rides. The neighbor is worried about safety because one of the rides looks particularly dangerous. She knows that you have taken physics and so asks you for advice. The ride in question has a \(4 \, \text{kg}\) chair which hangs freely from a \(10 \, \text{m}\) long chain attached to a pivot on the top of a tall tower. When the child enters the ride, the chain is hanging straight down. The child is then attached to the chair with a seat belt and shoulder harness. When the ride starts up, the chain rotates about the tower. Soon the chain reaches its maximum speed and remains rotating at that speed, which corresponds to one rotation about the tower every \(3 \, \text{seconds}\). When you ask the operator, he says the ride is perfectly safe. He demonstrates this by sitting in the stationary chair. The chain creaks but holds, and he weighs \(90 \, \text{kg}\).
Which of the following must be true for an object at translational equilibrium?
A person with a weight of \( 600 \) \( \text{N} \) stands on a scale in an elevator. What is the acceleration of the elevator when the scale reads \( 900 \) \( \text{N} \)?
Three blocks of masses \(5 \, \text{kg}\), \(4 \, \text{kg}\), and \(3 \, \text{kg}\) are placed side by side in that order. A \(25 \, \text{N}\) force applied on the \(5 \, \text{kg}\) block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the \(4 \, \text{kg}\) block exerts on the \(3 \, \text{kg}\) block.
Which pulls harder gravitationally, the Earth on the Moon, or the Moon on the Earth? Which accelerates more?
The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of \(0.80 \, \text{m}\) above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is \(0.35 \, \text{m}\). The mass of the marble is \(0.050 \, \text{kg}\). Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.
\(v = \sqrt{Rg\tan\theta}\)
\(10.3\,\text{m/s}\)
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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