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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."
A person is trying to judge whether a picture (mass = 1.42 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.62. What is the minimum amount of pressing force that must be used?
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
A 1kg and unknown mass (M) hangs on opposite sides of the pulley suspended from the ceiling. When the masses are released, M accelerates down at 5 m/s². What is M?
Two students push a 1750 kg car with a force of 758 N along a perfectly level road at a constant velocity of 4.00 m/s. Find the force of friction.
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
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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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