| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_{g\parallel}=mg\sin\theta\] | The component of gravitational force acting down the plane is \(mg\sin\theta\). |
| 2 | \[N=mg\cos\theta\] | The normal force \(N\) equals the perpendicular component of weight. |
| 3 | \[F_f=\mu N=\mu mg\cos\theta\] | Kinetic friction acts up the incline and has magnitude \(\mu N\). |
| 4 | \[F_{\text{net}}=mg\sin\theta-\mu mg\cos\theta\] | Net force along incline is downwards component minus friction. |
| 5 | \[F_{\text{net}}=ma\] | Apply Newton’s second law along the incline. |
| 6 | \[a=g(\sin\theta-\mu\cos\theta)\] | Solve for acceleration \(a\) after canceling \(m\). |
| 7 | \[a=9.8(0.5-0.20\times0.866)\] | Substitute \(g=9.8\,\text{m/s}^2\), \(\sin30^\circ=0.5\), and \(\cos30^\circ\approx0.866\). |
| 8 | \[\boxed{a\approx3.2\;\text{m/s}^2}\] | Compute the expression to find the block’s acceleration. |
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A 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 ° angle. The coefficient of kinetic friction of wood µk = 0.200. What magnitude of force should you apply to cause the box to slide down at a constant speed?
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?
The escape speed of an object of mass \( m \) from a planet of mass \( M \) and radius \( r \) depends on the gravitational constant and
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
The figure above shows a cart of mass \( M \) accelerating to the right with acceleration \( a \). A block of mass \( m \) is pressed against the cart’s front vertical surface and is held there only by friction. The coefficient of friction between the block and the cart is \( \mu \). What is the minimum acceleration \( a \) of the cart such that the block will not fall?
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.
A skydiver reaches a terminal velocity of \(55.0\, \mathrm{m/s}\). At terminal velocity, the skydiver no longer accelerates. The mass of the skydiver and her equipment is \(87.0\, \mathrm{kg}\). What is the force of friction acting on her?
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
A westward–moving car is changing its speed. The net force on the car ____.
A truck of mass 3500 kg hits the back of a small car of mass 1400 kg. Which car exerted more force on the other and why?
\(3.2\,\text{m/s}^2\)
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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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