| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ F_{g_x} = m g \sin \theta \] | Calculate the component of gravitational force along the incline for one block. |
| 2 | \[ F_{f_A} = \mu_A m_A g \cos \theta \] | Calculate the frictional force for block A. Frictional force is the product of the friction coefficient, mass, gravitational acceleration, and the cosine of the incline angle. |
| 3 | \[F_{f_B} = \mu_B m_B g \cos \theta \] | Calculate the frictional force for block B using its coefficient of friction. |
| 4 | \[ F_{\text{net}} = 2 F_{g_x} – F_{f_A} – F_{f_B} \] | Net force is the sum of both gravitational components minus the frictional forces for both blocks. |
| 5 | \[ F_{\text{net}} = 2 (m g \sin \theta) – (\mu_A m g \cos \theta \] \[+ \mu_B m g \cos \theta) \] |
Substitute the expressions for gravitational and frictional forces into the net force equation. |
| 6 | \[ F_{\text{net}} = 2 \cdot 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32) \]
\[+ 0.3 \cdot 5 \cdot 9.8 \cos(32)) \] |
Substitute known values: \( m = 5 \ \text{kg}, \ g = 9.8 \ \text{m/s}^2, \ \theta = 32^\circ \). |
| 7 | \[ F_{\text{net}} = 52.012 – 8.3 – 12.5 = \]
\[31.2 \ \text{N} \] |
Calculate the net force acting on both blocks by subtracting the calculated frictional forces from the gravitational force component. |
| 8 | \[a = \frac{F_{\text{net}}}{2m} = \frac{31.2}{2 \times 5} \] | Use Newton’s second law \( F = ma \) to solve for acceleration. |
| 9 | \[ \boxed{3.1 \ \text{m/s}^2} \] | Calculate the acceleration of the blocks. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ T = m_A g \sin \theta – F_{f_A} – m_A a \] | Consider block A. The tension is the net force after subtracting the force needed for acceleration and friction. |
| 2 | \[ T = 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32)) – 5 \cdot 3.1 \] | Substitute known values for mass, gravitational acceleration, incline angle, friction coefficient, and acceleration. |
| 3 | \[T = 26 – 8.3 – 15.6 = 2.1 \ \text{N} \] | Calculate the tension in the cord between the blocks. |
| 4 | \[ \boxed{2.1 \ \text{N}} \] | Express the tension in the cord as the final answer. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward?
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?
A brick slides on a horizontal surface. Which of the following will increase the magnitude of the frictional force on it?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.
A box is sliding down an incline at a constant speed of \( 2 \) \( \text{m s}^{-1} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \) \( \text{N} \). What is the force of gravity acting on the box?
According to Newton’s third law, each team in a tug of war pulls with equal force on the other team. What, then, determines which team will win?
A \(1509 \, \text{g}\) wood block is being pulled by the force meter at a constant velocity. Using the graph above, find:
If you stand on a bathroom scale in a moving elevator, does its reading change?
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
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
