| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = \frac{1}{2} a t^2\] | Use constant–acceleration kinematics with initial velocity \(v_i = 0\). |
| 2 | \[a = \frac{2\Delta x}{t^2}\] | Algebraically isolate \(a\). |
| 3 | \[a = \frac{2(10)}{(4.5)^2} = 0.99\,\text{m/s}^2\] | Insert \(\Delta x = 10\,\text{m}\) and \(t = 4.5\,\text{s}\) to find \(a \approx 0.99\,\text{m/s}^2\). |
| 4 | \[F_{1x}=100\cos20^\circ,\;F_{2x}=-400\cos40^\circ,\;F_{3x}=500\cos10^\circ\] | Resolve each force into horizontal components; left is negative, right positive. |
| 5 | \[F_{1x}=93.97\,\text{N},\;F_{2x}=-306.6\,\text{N},\;F_{3x}=492.4\,\text{N}\] | Evaluate the trigonometric products. |
| 6 | \[F_{x,\text{applied}} = 279.8\,\text{N}\] | Sum \(F_{1x}+F_{2x}+F_{3x}\) to get the net applied horizontal force. |
| 7 | \[F_{1y}=+100\sin20^\circ,\;F_{2y}=-400\sin40^\circ,\;F_{3y}=-500\sin10^\circ\] | Resolve forces into vertical components; upward is positive. |
| 8 | \[F_{1y}=+34.2\,\text{N},\;F_{2y}=-257.1\,\text{N},\;F_{3y}=-86.8\,\text{N}\] | Compute the numeric values. |
| 9 | \[F_{y,\text{applied}} = -309.7\,\text{N}\] | Add the vertical components to find a downward net external load of \(309.7\,\text{N}\). |
| 10 | \[N = mg + 309.7\,\text{N}\] | Vertical equilibrium requires \(N + F_{1y} = mg + |F_{2y}| + |F_{3y}|\); simplify to this expression for the normal force. |
| 11 | \[f_k = \mu_k N = 0.2\,(mg + 309.7)\] | Kinetic friction opposes motion with magnitude \(\mu_k N\). |
| 12 | \[F_{\text{net}} = F_{x,\text{applied}} – f_k\] | Net horizontal force equals applied force minus friction (leftward). |
| 13 | \[ma = 279.8 – 0.2(mg + 309.7)\] | Apply Newton’s second law \(\sum F_x = ma\). |
| 14 | \[m(a + 1.96) = 217.8\] | Use \(g = 9.8\,\text{m/s}^2\) and simplify: \(0.2g = 1.96\,\text{N/kg}\). |
| 15 | \[m = \frac{217.8}{a + 1.96}\] | Isolate the unknown mass. |
| 16 | \[m = \frac{217.8}{0.99 + 1.96} = 73.9\,\text{kg}\] | Insert \(a \approx 0.99\,\text{m/s}^2\) to calculate \(m\). |
| 17 | \[\boxed{m \approx 7.4 \times 10^{1}\,\text{kg}}\] | Express the mass to two significant figures. |
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.
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
Will Clark throws a baseball with a horizontal component of velocity of \(25 \, \text{m/s}\). It takes \(3 \, \text{s}\) to come back to its original height. Calculate the baseball’s:
A ball is launched horizontally from a height. At the same time, another ball is dropped vertically from the same height. Which hits the ground first?
An object is moving at constant velocity. Which of the following could be the free-body diagram representing the forces acting on the object?

Three blocks of masses \(m_3 = 1.0 \, \text{kg}\), \(m_2 = 2.0 \, \text{kg}\), and \(m_1 = 4.0 \, \text{kg}\) are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.
If an elephant were chasing you, its enormous mass would be most threatening. But if you zigzagged, its mass would be to your advantage. Why?
The displacement \(x\) of an object moving in one dimension is shown above as a function of time \(t\). The velocity of this object must be
Three blocks are stacked on top of one another. The top block has a mass of \( 4.6 \, \text{kg} \), the middle one has a mass of \( 1.2 \, \text{kg} \), and the bottom one has a mass of \( 3.7 \, \text{kg} \).
Identify and calculate any normal forces between the objects.
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
An object is released from rest near the surface of a planet. The velocity of the object as a function of time is expressed in the following equation. \( v_y = (-3) \, \text{m/s}^2 \, t \) All frictional forces are considered to be negligible. What distance does the object fall \( 10 \) \( \text{s} \) after it is released from rest?
\(74\,\text{kg}\)
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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
NEW! PHY AI accurately solves all questions
🔥 Get up to 30% off Elite Physics Tutoring
🧠 NEW! Learn Physics From Scratch Self Paced Course
🎯 Need exam style practice questions?