0 attempts
0% avg
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m_{\text{top}} = 20\,\text{kg} + 30\,\text{kg} = 50\,\text{kg}\] | Combine the masses of the top two blocks (20 kg and 30 kg) that are above the contact between the 30 kg and 40 kg blocks. |
| 2 | \[50\,\text{kg}\,g – F = 50\,\text{kg}\,a\] | Apply Newton’s second law to the combined system. The total weight of the top blocks (\(50\,\text{kg}\,g\)) minus the contact force \(F\) (exerted by the 40 kg block upward on the 30 kg block) must equal the net force needed to accelerate the system downward at \(a = 3.2\,\text{m/s}^2\). |
| 3 | \[F = 50\,\text{kg}\,(g – a)\] | Solve for the contact force \(F\) by isolating it in the equation. |
| 4 | \[F = 50\,\text{kg}\,(9.8\,\text{m/s}^2 – 3.2\,\text{m/s}^2) = 50\,\text{kg}\,(6.6\,\text{m/s}^2) = 330\,\text{N}\] | Substitute \(g = 9.8\,\text{m/s}^2\) and \(a = 3.2\,\text{m/s}^2\) into the equation and calculate the force. |
| 5 | \[\boxed{330\,\text{N}}\] | This is the contact force between the 30 kg and 40 kg masses. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assume the pulley has a frictional force of \(5.7\) Newtons acting on the outer edge of the pulley.
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
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.
Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle \( \theta = 32^\circ \) above the horizontal. If the coefficients of friction are \( \mu_A = 0.2 \) and \( \mu_B = 0.3 \) and if \( m_A = m_B = 5.0 \) \( \text{kg} \), determine:
When a horizontal force of \( 4.5 \, \text{N} \) acts on a block on a resistance-free surface, it produces an acceleration of \( 2.5 \, \text{m/s}^2 \). Suppose a second \( 4.0 \, \text{kg} \) block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block.
A \(1 \, \text{kg}\) mass and an unknown mass \(M\) hang on opposite sides of a pulley suspended from the ceiling. When the masses are released, \(M\) accelerates downward at \(5 \, \text{m/s}^2\). Find the value of \(M\).
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
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.
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?
A train consists of \(50\) cars, each of which has a mass of \(6.1 \times 10^{3} \, \text{kg}\). The train has an acceleration of \(8.0 \times 10^{-2} \, \text{m/s}^2\). Ignore friction and determine the tension in the coupling at the following places:
\(330 \, \text{N}\)
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
