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Part a: Calculate the acceleration of the system
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{net,A}} = m_A \cdot a [/katex] | Net force on mass A equals mass times acceleration. |
| 2 | [katex] F_{\text{net,B}} = m_B \cdot a [/katex] | Net force on mass B equals mass times acceleration. |
| 3 | [katex] F_{\text{net,A}} = T – m_A \cdot g [/katex] | Tension upwards minus weight of A downwards. |
| 4 | [katex] F_{\text{net,B}} = m_B \cdot g – T [/katex] | Weight of B downwards minus tension upwards. |
| 5 | [katex] m_A \cdot a = T – m_A \cdot g [/katex] | Substitute step 1 into step 3. |
| 6 | [katex] m_B \cdot a = m_B \cdot g – T [/katex] | Substitute step 2 into step 4. |
| 7 | [katex] m_A \cdot a + m_B \cdot a = m_B \cdot g – m_A \cdot g [/katex] | Add step 5 and step 6 equations. |
| 8 | [katex] a = \frac{(m_B – m_A) \cdot g}{m_A + m_B} [/katex] | Solve for acceleration . |
Use the given number from the problem.
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 9 | [katex] a = \frac{(2.4 – 3.2) \cdot 9.8}{3.2 + 2.4} [/katex] | Plug in known values. |
| 10 | [katex] a = \frac{-0.8 \cdot 9.8}{5.6} [/katex] | Simplify the numerator and denominator. |
| 11 | [katex] a = -1.4 , \text{m/s}^2 [/katex] | Calculate the acceleration. |
[katex] \boxed{a = -1.4 , \text{m/s}^2} [/katex] Negative sign indicates downward direction.
Part b: Calculate the tension in the string
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] T = m_A \cdot (g + a) [/katex] | Tension equals mass A times (gravity plus acceleration). |
Using the previously calculated acceleration:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 2 | [katex] T = 3.2 \cdot (9.8 – 1.4) [/katex] | Plug in known values for mass and acceleration |
| 3 | [katex] T = 3.2 \cdot 8.4 [/katex] | Simplify the terms in the parentheses. |
| 4 | [katex] T = 26.88 , \text{N} [/katex] | Calculate the tension. |
[katex] \boxed{T = 26.88 , \text{N}} [/katex]
Part c: Calculate the final speed of mass A before it hits the ground
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] v = \sqrt{2 \cdot} [/katex] | a |
| 2 | [katex] v = \sqrt{2 \cdot 1.4 \cdot 0.5} [/katex] | Plug in the magnitude of and . |
| 3 | [katex] v = \sqrt{1.4} [/katex] | Calculate the expression under the square root. |
| 4 | [katex] v = 1.18 , \text{m/s} [/katex] | Find the square root to get the final velocity. |
[katex] \boxed{v = 1.18 , \text{m/s}} [/katex]
Just ask: "Help me solve this problem."
Why do pilots sometimes black out while pulling out at the bottom of a dive?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assume the pulley has a frictional force of \(5.7\) Newtons.
If the acceleration of an object is \( 0 \), are no forces acting on it? Explain.
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
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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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