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| 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. |
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A \( 1 \) \( \text{kg} \) mass on a \( 37^{\circ} \) incline is connected to a \( 3.0 \) \( \text{kg} \) mass on a horizontal surface, as shown. The surfaces and the pulley are frictionless. If \( F = 12 \) \( \text{N} \):
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A group of astronauts in a spaceship are attempting to land on Mars. As they approach the planet, they begin to plan their descent to the surface.
In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:
Above is the graph of the velocity vs. time of a duck flying due south for the winter. At what point might the duck begin reversing directions?
If the acceleration of an object is \( 0 \), are no forces acting on it? Explain.
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball \( B \) passes ball \( A \):
A student is running at her top speed of \( 5.0 \, \text{m/s} \) to catch a bus, which is stopped at the bus stop. When the student is still \( 40.0 \, \text{m} \) from the bus, it starts to pull away, moving with a constant acceleration of \( 0.170 \, \text{m/s}^2 \).
A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of \( 1500 \, \text{m} \). The rocket is launched with an initial velocity of \( 180 \, \text{m/s} \). Find the following:
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.
\(74\,\text{kg}\)
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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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