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Objective: Calculate the acceleration of the two objects and the tension in the string.
For Acceleration
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]\text{Force on heavier mass, } F_1 = m_1g[/katex] | Weight of the heavier object (49.0 kg). |
| 2 | [katex]\text{Force on lighter mass, } F_2 = m_2g[/katex] | Weight of the lighter object (24.0 kg). |
| 3 | [katex]\text{Net force, } F_{\text{net}} = F_1 – F_2[/katex] | The difference in weights provides the net force. |
| 4 | [katex]F_{\text{net}} = m_1g – m_2g[/katex] | Substitute the values of [katex]F_1[/katex] and [katex]F_2[/katex]. |
| 5 | [katex]F_{\text{net}} = (m_1 – m_2)g[/katex] | Factor out [katex]g[/katex]. |
| 6 | [katex]a = \frac{F_{\text{net}}}{m_1 + m_2}[/katex] | Newton’s second law, acceleration equals net force divided by total mass. |
| 7 | [katex]a = \frac{(m_1 – m_2)g}{m_1 + m_2}[/katex] | Combine steps 5 and 6. |
| 8 | [katex]a = \frac{(49.0\text{ kg} – 24.0\text{ kg})(9.8\text{ m/s}^2)}{49.0\text{ kg} + 24.0\text{ kg}}[/katex] | Substitute the masses and gravitational acceleration. |
| 9 | [katex]a = \frac{25.0\text{ kg} \times 9.8\text{ m/s}^2}{73.0\text{ kg}}[/katex] | Simplify the equation. |
| 10 | [katex]a = 3.356\text{ m/s}^2[/katex] | Calculate to find acceleration. |
Final answer for acceleration: [katex]\boxed{a = 3.356\text{ m/s}^2}[/katex]
For Tension in the String
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]T – m_2g = m_2a[/katex] | Newton’s second law for the lighter object. |
| 2 | [katex]T = m_2a + m_2g[/katex] | Rearrange to solve for tension, [katex]T[/katex]. |
| 3 | [katex]T = m_2(a + g)[/katex] | Factor out [katex]m_2[/katex]. |
| 4 | [katex]T = 24.0\text{ kg}(3.356\text{ m/s}^2 + 9.8\text{ m/s}^2)[/katex] | Substitute the mass of the lighter object and calculated acceleration. |
| 5 | [katex]T = 24.0\text{ kg} \times 13.156\text{ m/s}^2[/katex] | Add [katex]a[/katex] and [katex]g[/katex]. |
| 6 | [katex]T = 315.744\text{ N}[/katex] | Calculate to find tension. |
Final answer for tension: [katex]\boxed{T = 315.744\text{ N}}[/katex]
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A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
An object is moving at constant velocity. Which of the following could be the free-body diagram representing the forces acting on the object?
3.36 m/s2, T = 316 N
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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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