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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ F_{g_x} = m g \sin \theta \] | Calculate the component of gravitational force along the incline for one block. |
| 2 | \[ F_{f_A} = \mu_A m_A g \cos \theta \] | Calculate the frictional force for block A. Frictional force is the product of the friction coefficient, mass, gravitational acceleration, and the cosine of the incline angle. |
| 3 | \[F_{f_B} = \mu_B m_B g \cos \theta \] | Calculate the frictional force for block B using its coefficient of friction. |
| 4 | \[ F_{\text{net}} = 2 F_{g_x} – F_{f_A} – F_{f_B} \] | Net force is the sum of both gravitational components minus the frictional forces for both blocks. |
| 5 | \[ F_{\text{net}} = 2 (m g \sin \theta) – (\mu_A m g \cos \theta \] \[+ \mu_B m g \cos \theta) \] |
Substitute the expressions for gravitational and frictional forces into the net force equation. |
| 6 | \[ F_{\text{net}} =Â 2 \cdot 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32) \]
\[+ 0.3 \cdot 5 \cdot 9.8 \cos(32)) \] |
Substitute known values: \( m = 5 \ \text{kg}, \ g = 9.8 \ \text{m/s}^2, \ \theta = 32^\circ \). |
| 7 | \[ F_{\text{net}} = 52.012 – 8.3 – 12.5 = \]
\[31.2 \ \text{N} \] |
Calculate the net force acting on both blocks by subtracting the calculated frictional forces from the gravitational force component. |
| 8 | \[a = \frac{F_{\text{net}}}{2m} = \frac{31.2}{2 \times 5} \] | Use Newton’s second law \( F = ma \) to solve for acceleration. |
| 9 | \[ \boxed{3.1 \ \text{m/s}^2} \] | Calculate the acceleration of the blocks. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ T = m_A g \sin \theta – F_{f_A} – m_A a \] | Consider block A. The tension is the net force after subtracting the force needed for acceleration and friction. |
| 2 | \[ T = 5 \cdot 9.8 \sin(32) – (0.2 \cdot 5 \cdot 9.8 \cos(32)) – 5 \cdot 3.1 \] | Substitute known values for mass, gravitational acceleration, incline angle, friction coefficient, and acceleration. |
| 3 | \[T = 26 – 8.3 – 15.6 = 2.1 \ \text{N} \] | Calculate the tension in the cord between the blocks. |
| 4 | \[ \boxed{2.1 \ \text{N}} \] | Express the tension in the cord as the final answer. |
Just ask: "Help me solve this problem."
A 25.0-kg box is released on a 23.5° incline and accelerates down the incline at 0.35 m/s2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?
A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be [katex]6 \times 10^{24} \text{ kg}[/katex] the the radius of Earth to be [katex]6.4 \times 10^6 \text{ m}[/katex]. What is the satellite’s velocity?
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?
A car can decelerate at \( -3.80 \, \text{m/s}^2 \) without skidding when coming to rest on a level road. What would its deceleration be if the road is inclined at \( 9.3^\circ \) and the car moves uphill? Assume the same static friction coefficient.
A rope of negligible mass supports a block that weighs \(30 N\), as shown above. The breaking strength of the rope is \( 50 N\). The largest acceleration that can be given to the block by pulling up on it with the rope without breaking the rope is most nearly
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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