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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. |
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A child on Earth has a weight of \(500 \, \text{N}\). Determine the weight of the child if the Earth were to triple in both mass and radius (\(3M\) and \(3r\)).
Two closed containers look the same, but one is packed with lead and the other with a few feathers. How could you determine which has more mass if you and the containers were orbiting in a weightless condition in outer space?
Two objects attract each other gravitationally with a force of \( 2.5 \times 10^{-10} \) \( \text{N} \) when they are \( 0.25 \) \( \text{m} \) apart. Their total mass is \( 4.00 \) \( \text{kg} \). Find their individual masses.
Why do pilots sometimes black out while pulling out at the bottom of a dive?
Two students push a \(1750\, \mathrm{kg}\) car with a force of \(758\, \mathrm{N}\) along a perfectly level road at a constant velocity of \(4.00\, \mathrm{m/s}\). Find the force of friction.
A person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?
A pair of fuzzy dice is hanging by a string from your rearview mirror. You speed up from a stoplight. During the acceleration, the dice do not move vertically; the string makes an angle of \( 22^\circ \) with the vertical. The dice have a mass of \( 0.10 \, \text{kg} \). Determine the acceleration.
A \(0.5 \, \text{mm}\) wire made of carbon and manganese can just barely support the weight of a \(70.0 \, \text{kg}\) person that is holding on vertically. Suppose this wire is used to lift a \(45.0 \, \text{kg}\) load. What maximum vertical acceleration can be achieved without breaking the wire?
A \(3300 \, \text{m}\)-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The Earth’s radius is \(6400 \, \text{km}\) and the Earth’s mass is \(5.97 \times 10^{24} \, \text{kg}\).
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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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