| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_{\text{push}} = -F_{\text{back}}\] | By Newtons Third Law, the force you exert on the box (\(F_{\text{push}}\)) and the force the box exerts on you (\(F_{\text{back}}\)) are equal in magnitude and opposite in direction. |
| 2 | \[\vec{F}_{\text{net}} = m\vec{a}\] | The box accelerates because the net force on it (push minus friction) is non Zero, but this net force concept does not alter the action reaction pairs equality from Step 1. |
| 3 | \[|F_{\text{push}}| = |F_{\text{back}}|\] | Magnitude equality follows directly from Step 1, independent of the boxs mass or its acceleration. |
| 4 | \[\boxed{\text{(d)}}\] | Thus the correct choice is (d): the force you exert on the box equals the force the box exerts on you. |
| 5 | \[\text{(a)}\] | Incorrect: Newtons Third Law requires a reaction force; the box must push back on you. |
| 6 | \[\text{(b)}\] | Incorrect: You clearly push on the box; omitting this breaks the action reaction pair. |
| 7 | \[\text{(c)}\] | Incorrect: The forces cannot differ in magnitude; they are an action reaction pair, not a net force scenario. |
| 8 | \[\text{(e)}\] | Incorrect for the same reason as (c); the magnitudes must be equal. |
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.
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only \( 0.75 \) of her regular weight. Calculate the acceleration of the elevator, and find the direction of acceleration.
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 traffic light hangs from a pole as shown in the diagram. The uniform aluminum pole AB is of length \( 7.20 \) \( \text{m} \) and has a mass of \( 12.0 \) \( \text{kg} \). The mass of the traffic light is \( 21.5 \) \( \text{kg} \). The point C is located \( 3.80 \) \( \text{m} \) vertically above the pivot A. A massless horizontal cable CD is attached at C and connects to the pole at point D, where the pole makes an angle of \( 37^{\circ} \) with the cable.
A \( 1.5 \) \( \text{kg} \) block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is \( 0.300 \) and the coefficient of static friction is \( 0.400 \), calculate the amount of time that passes from when the force is applied to when the block stops.
A car is going over the top of a hill whose curvature approximates a circle of radius \( 350 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 10\% \) less than their normal weight?
When a falling meteoroid is at a distance above the Earth’s surface of \( 3.00 \) times the Earth’s radius, what is its acceleration due to the Earth’s gravitation?
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
Two spherical objects have equal masses and experience a gravitational force of \( 25 \) \( \text{N} \) towards one another. Their centers are \( 36 \) \( \text{cm} \) apart. Determine each of their masses.
A ball falls straight down through the air under the influence of gravity. There is a retarding force \(F\) on the ball with magnitude given by \(F=bv\), where \(v\) is the speed of the ball and \(b\) is a positive constant. The ball reaches a terminal velocity after a time \(t\). The magnitude of the acceleration at time \(t/2\) is
The coefficient of static friction between hard rubber and normal street pavement is about \(0.85\). On how steep a hill (maximum angle) can you leave a car parked?
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?