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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( F_{\text{net}} = m_{4}a \) | The net force acting on the \(4 \, \text{kg}\) block is equal to its mass times its acceleration. |
| 2 | \( F_{\text{gravity}} = m_{4}g \) | The force of gravity acting on the \(4 \, \text{kg}\) block is equal to its mass times the acceleration due to gravity \(g\). |
| 3 | \( F_{\text{net}} = m_{4}g – T \) | The net force acting on the \(4 \, \text{kg}\) block is the gravitational force minus the tension in the string. |
| 4 | \( m_{4}a = m_{4}g – T \) | Substitute the net force into Newton’s second law. |
| 5 | \( T = m_{1}g + m_{1}a + m_{2}g + m_{2}a \) | The tension in the string also depends on the forces acting on the \(1.0\, \text{kg}\) and \(2.0 \, \text{kg}\) blocks. |
| 6 | \( T = m_{1}(g + a) + m_{2}(g + a) \) | Combine the tensions for the \(1.0 \, \text{kg}\) and \(2.0 \, \text{kg}\) blocks since they share the same strings. |
| 7 | \( m_{4}a = m_{4}g – \left[m_{1}(g + a) + m_{2}(g + a)\right] \) | Substitute the tension \(T\) from step 6 into the equation from step 4. |
| 8 | \( 4a = 4g – (1 + 2)(g + a) \) | Substitute \(m_{4} = 4\, \text{kg}\), \(m_{1} = 1\, \text{kg}\), and \(m_{2} = 2\, \text{kg}\). |
| 9 | \( 4a = 4g – 3(g + a) \) | Combine the masses for the \(1.0 \, \text{kg}\) and \(2.0 \, \text{kg}\) blocks. |
| 10 | \( 4a = 4g – 3g -3a \) | Distribute the 3 to both terms. |
| 11 | \( 7a = g \) | Combine like terms to isolate \(a\). |
| 12 | \( a = \frac{g}{7} \approx \frac{9.8\, \text{m/s}^2}{7} \approx 1.4\, \text{m/s}^2 \) | Solve for \(a\), the acceleration of the \(4 \, \text{kg}\) block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(T = m_4 g – m_4 a\) | The tension in the string is the gravitational force on the \(4 \, \text{kg}\) block minus the force due to its acceleration. |
| 2 | \(T = 4 \times 9.8 – 4 \times 1.4\) | Substitute \(m_4 = 4\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\), and \(a = 1.4\, \text{m/s}^2\). |
| 3 | \(T = 39.2 – 5.6\) | Calculate the products. |
| 4 | \( T = 33.6 \, \text{N} \) | Final value for the tension in the string supporting the \(4 \, \text{kg}\) block. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_1 = m_1 (g + a) \) | The tension in the string is the gravitational force on the \(1 \, \text{kg}\) block plus the force due to its acceleration. |
| 2 | \( T_1 = 1 \times (9.8 + 1.4) \) | Substitute \(m_1 = 1\, \text{kg}\), \(g = 9.8\, \text{m/s}^2\), and \(a = 1.4\, \text{m/s}^2\). |
| 3 | \( T_1 = 1 \times 11.2 \) | Combine the terms inside the parentheses. |
| 4 | \( T_1 = 11.2 \, \text{N} \) | Final value for the tension in the string connected to the \(1 \, \text{kg}\) block. |
Just ask: "Help me solve this problem."
A 0.5 mm wire made of carbon and manganese can just barely support the weight of a 70.0 kg person that is holding on vertically. Suppose this wire is used to lift a 45.0 kg load. What maximum vertical acceleration can be achieved without breaking the wire?
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 rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?
Four blocks of masses \( 20 \, \text{kg}, \, 30 \, \text{kg}, \, 40 \, \text{kg}, \, \text{and} \, 50 \, \text{kg} \) are stacked on top of one another in an elevator in order of decreasing mass with the lightest mass on the top of the stack. The elevator moves downward with an acceleration of \( 3.2 \, \text{m/s}^2 \). Find the contact force between the \( 30 \, \text{kg} \) and \( 40 \, \text{kg} \) masses.
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?
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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] |
General Metric Conversion Chart
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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