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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 pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
A 100 kg person is riding a 10 kg bicycle up a 25° hill. The hill is long and the coefficient of static friction is 0.9. The person rides 10 m up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of 7 m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
A sled glides across ice and eventually stops. This stopping is best explained by ____.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?
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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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