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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]
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A coffee cup on the dashboard of a car slides forward when the driver decelerates from \(45 ~ \frac{\text{km}}{\text{hr}}\) to rest in \(3.5 \, \text{s}\) or less. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are completely horizontal.
A box rests on the (frictionless) bed of a truck. The truck driver starts the truck and accelerates forward. The box immediately starts to slide toward the rear of the truck bed.
A spherical balloon of mass \( 226 \) \( \text{kg} \) is filled with helium gas until its volume is \( 325 \) \( \text{m}^3 \). Assume the density of air is \( 1.29 \) \( \text{kg/m}^3 \) and the density of helium is \( 0.179 \) \( \text{kg/m}^3 \).
Two balls have their centers \( 2.0 \) \( \text{m} \) apart. One ball has a mass of \( 8.0 \) \( \text{kg} \). The other has a mass of \( 6.0 \) \( \text{kg} \). What is the gravitational force between them?
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?
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
A \( 1 \) \( \text{kg} \) mass on a \( 37^{\circ} \) incline is connected to a \( 3.0 \) \( \text{kg} \) mass on a horizontal surface, as shown. The surfaces and the pulley are frictionless. If \( F = 12 \) \( \text{N} \):
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 \(5.97 \times 10^{24} \, \text{kg}\).
Two ice skaters push off against each other. Skater A has twice the mass of Skater B. Which statement is correct?
A block rests on a flat plane inclined at an angle of \(30^\circ\) with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
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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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