0 attempts
0% avg
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ K_{\text{trans}} = \frac{1}{2} M_{\text{tot}} v^2 \] | All three wagons have the same total mass and are accelerated to the same speed, so the translational kinetic energy is identical for each. |
| 2 | \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \quad \text{with} \quad \omega = \frac{v}{R} \] | This is the rotational kinetic energy for a wheel rolling without slipping. |
| 3 | \[ I_{\text{disk}} = \frac{1}{2} M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,disk}} = \frac{1}{2}\left(\frac{1}{2} M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4} M_{w} v^2 \] | For a solid disk wheel (used in Wagons A and B), the moment of inertia is substituted to find its rotational kinetic energy. |
| 4 | \[ I_{\text{hoop}} = M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,hoop}} = \frac{1}{2}\left(M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2} M_{w} v^2 \] | For a hollow hoop wheel (used in Wagon C), the moment of inertia is larger, leading to greater rotational energy. |
| 5 | \[ K_{\text{total, wheel}} = K_{\text{trans, wheel}} + K_{\text{rot}} = \frac{1}{2} M_{w} v^2 + K_{\text{rot}} \] | Each wheel has translational kinetic energy (as it moves with the wagon) plus its rotational kinetic energy. |
| 6 | For a disk: \[ K_{\text{total, disk}} = \frac{1}{2} M_{w} v^2 + \frac{1}{4} M_{w} v^2 = \frac{3}{4} M_{w} v^2 \] |
This is the effective energy per disk wheel. If the wheel’s mass were a simple point mass, it would only have \(\frac{1}{2}M_{w}v^2\); thus, the extra energy due to rotation is \(\Delta K_{\text{disk}} = \frac{3}{4} M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{4} M_{w} v^2\). |
| 7 | For a hoop: \[ K_{\text{total, hoop}} = \frac{1}{2} M_{w} v^2 + \frac{1}{2} M_{w} v^2 = M_{w} v^2 \] |
Similarly, the extra energy due to rotation for a hoop is \(\Delta K_{\text{hoop}} = M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{2} M_{w} v^2\), which is larger than that for a disk of the same mass. |
| 8 | Wagon A (disk, \(M_{w}=0.5\,\text{kg}\)): \[ \Delta K_{A} = 4\left(\frac{1}{4}\times 0.5\, v^2\right) = 4(0.125\, v^2) = 0.5\, v^2 \]Wagon B (disk, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{B} = 4\left(\frac{1}{4}\times 0.2\, v^2\right) = 4(0.05\, v^2) = 0.2\, v^2 \]Wagon C (hoop, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{C} = 4\left(\frac{1}{2}\times 0.2\, v^2\right) = 4(0.1\, v^2) = 0.4\, v^2 \] |
Each wagon has four wheels. The extra energy due to wheel rotation is calculated by multiplying the additional energy per wheel by four. |
| 9 | With \(v = 10\,\text{m/s}\) (\(v^2 = 100\)): \[ \Delta K_{A} = 0.5 \times 100 = 50\, \text{J} \] \[ \Delta K_{B} = 0.2 \times 100 = 20\, \text{J} \] \[ \Delta K_{C} = 0.4 \times 100 = 40\, \text{J} \] |
Substitute the final speed into each expression to obtain the numerical extra energy that must be supplied to accelerate the wheels. |
| 10 | \[\boxed{\text{Wagon A}}\] | Since the translational energy is the same for all wagons, the wagon with the greatest extra wheel energy (50 J) requires the most energy input. Therefore, Wagon A requires the most energy. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
A uniform ladder with mass \( m_2 \) and length \( L \) rests against a smooth wall. A do-it-yourself enthusiast of mass \( m_1 \) stands on the ladder a distance \( d \) from the bottom (measured along the ladder). The ladder makes an angle \( \theta \) with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude \( f \) between the floor and the ladder. \( N_1 \) is the magnitude of the normal force exerted by the wall on the ladder, and \( N_2 \) is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first?
A windmill blade with a rotational inertia of \( 6.0 \) \( \text{kg} \cdot \text{m}^2 \) has an initial angular velocity of \( 8 \) \( \text{rad/s} \) in the clockwise direction. It is then given an angular acceleration of \( 4 \) \( \text{rad/s}^2 \) in the clockwise direction for \( 10 \) seconds. What is the change in rotational kinetic energy of the blade over this time interval?
A uniform rod of mass \( M_0 \) and length \( L \) is free to rotate about a pivot at its left end and is released from rest when the rod is \( 30^{\circ} \) below the horizontal, as shown in the figure. With respect to the pivot, the rod has rotational inertia \( I_0 = \dfrac{1}{3} M_0 L^2 \). Which of the following expressions correctly represents the magnitude of the net torque exerted on the rod about the pivot at the moment the rod is released?
A solid sphere \( \left( I = \frac{2}{5}MR^2 \right) \) and a solid cylinder \( \left( I = \frac{1}{2}MR^2 \right) \), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
A friend is balancing a fork on one finger. Which of the following are correct explanations of how he accomplishes this? Select two answers.
Three forces of equal magnitude are applied to a \( 3 \)-m by \( 2 \)-m rectangle. Force \( F_1 \) and \( F_2 \) act at \( 45^\circ \) angles to the vertical as shown, while \( F_3 \) acts horizontally.
Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?
A solid sphere of mass [katex] 1.5 \, \text{kg} [/katex] and radius [katex] 15 \, \text{cm} [/katex] rolls without slipping down a [katex] 35^\circ[/katex] incline that is [katex] 7 \, \text{m} [/katex] long. Assume it started from rest. The moment of inertia of a sphere is [katex] I= \frac{2}{5}MR^2 [/katex].
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
