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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 45 kg crate accelerates at 1.65 m/s2 when pulled by a rope with a force of 200 N. Find the angle the rope is pulled at. Friction is negligible.
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle [katex] \theta [/katex] to the horizontal. If the coefficients of friction are µA = .2 and µB = .3 and if mA = mB = 5.0 kg determine:
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
Three blocks of masses 5, 4, and 3 kg are placed side by side in that order. A 25 N force applied on the 5 kg block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the 4 kg block exerts on the 3 kg block.
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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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