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| Derivation/Formula | Reasoning |
|---|---|
| \[m_{\text{tot}} = 5 + 4 + 3 = 12\,\text{kg}\] | The three blocks move together, so their masses add: \(m_{\text{tot}} = 12\,\text{kg}\). |
| \[a = \frac{F}{m_{\text{tot}}} = \frac{25}{12}\,\text{m/s}^2\] | Using Newton’s second law \(F = m a\), solve for acceleration: \(a = F / m_{\text{tot}}\). |
| \[\boxed{a = \frac{25}{12}\,\text{m/s}^2}\] | Acceleration of all three blocks. |
| Derivation/Formula | Reasoning |
|---|---|
| \[F_N = m_3 a\] | For the \(3\,\text{kg}\) block, the only horizontal force is the push from the \(4\,\text{kg}\) block, so \(F_N\) supplies its acceleration \(a\). |
| \[F_N = 3 \times \frac{25}{12} = \frac{75}{12}\,\text{N}\] | Substitute \(m_3 = 3\,\text{kg}\) and the previously found \(a = \frac{25}{12}\,\text{m/s}^2\). |
| \[\boxed{F_N = \frac{75}{12}\,\text{N}}\] | Normal force from the \(4\,\text{kg}\) block on the \(3\,\text{kg}\) block. |
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Suppose an object is accelerated by a force of \( 100 \) \( \text{N} \). Suddenly a second force of \( 100 \) \( \text{N} \) in the opposite direction is exerted on the object, so that the forces cancel. The object
A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?
What is the mass of a dog that weighs \(58 \, \text{N}\) on Earth?
A person stands on a scale in an elevator. If the scale reads \( 600 \, \text{N} \) when that person is riding upward at a constant velocity of \( 4 \, \text{m/s} \), what is the scale reading when the elevator is at rest? Hint: The reading on the scale is simply the normal force.
A \(10 \, \text{kg}\) box is pushed to the right by an unknown force at an angle of \(25^\circ\) below the horizontal while a friction force of \(50 \, \text{N}\) acts on the box as well. The box accelerates from rest and travels a distance of \(4 \, \text{m}\) where it is moving at \(3 \, \text{m/s}\).
What is the weight of a person who has a mass of \(75 \, \text{kg}\)?
Which of the following best explains why astronauts experience weightlessness while orbiting the earth?
Two blocks, A and B, are connected by a light string that passes over a frictionless pulley. Block A, of mass \( 10 \) \( \text{kg} \), rests on a rough plane that makes an angle of \( 45^{\circ} \) with the horizontal, while block B, of mass \( 17 \) \( \text{kg} \), hangs vertically. Starting from rest, what is the minimum coefficient of static friction between block A and the plane required to keep the system in static equilibrium?
A forward horizontal force of \(12 \, \text{N}\) is used to pull a \(240 \, \text{N}\) crate at constant velocity across a horizontal floor. The coefficient of friction is
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
\(\frac{25}{12}\,\text{m/s}^2\)
\(\frac{75}{12}\,\text{N}\)
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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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