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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Accelerarion of the particle when its displacement is 6 m | ||
| 1 | \[F = ma\] | Newton’s second law relates force \( F \), mass \( m \), and acceleration \( a \). |
| 2 | \[a = \frac{F}{m}\] | Rearrange the formula to solve for acceleration. |
| 3 | \[a = \frac{4\, \text{N}}{0.20\, \text{kg}}\] | Substitute the force from the graph (4 N) and the mass (0.20 kg). |
| 4 | \[a = 20\, \text{m/s}^2\] | Calculate the acceleration. |
| (b) Time taken for the object to be displaced the first 12 m | ||
| 1 | \[\Delta x = v_i t + \frac{1}{2} a t^2\] | Using the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[12 = \frac{1}{2} \cdot 20 \cdot t^2\] | Substitute \( \Delta x = 12 \) m and \( a = 20 \text{ m/s}^2 \). |
| 3 | \[12 = 10 t^2\] | Simplify the equation. |
| 4 | \[t^2 = 1.2\] | Divide both sides by 10. |
| 5 | \[t = \sqrt{1.2}\] | Solve for \( t \). |
| 6 | \[t \approx 1.095\, \text{s}\] | Calculate the time taken. |
| (c) The amount of work done by the net force in displacing the object the first 12 m | ||
| 1 | \[W = F \Delta x\] | Work done \( W \) is the product of force and displacement. |
| 2 | \[W = 4 \times 12\] | Substitute \( F = 4 \text{ N} \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[W = 48 \text{ J}\] | Calculate the work done. |
| (d) The speed of the object at displacement \( x = 12 \text{ m} \) | ||
| 1 | \[v_x^2 = v_i^2 + 2a \Delta x\] | Use the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[v_x^2 = 0 + 2 \cdot 20 \cdot 12\] | Substitute \( a = 20 \text{ m/s}^2 \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[v_x^2 = 480\] | Calculate \( v_x^2 \). |
| 4 | \[v_x = \sqrt{480}\] | Solve for \( v_x \). |
| 5 | \[v_x \approx 21.9 \, \text{m/s}\] | Calculate the velocity. |
| (e) The final speed of the object at displacement \( x = 20 \text{ m} \) | ||
| 1 | \[W_{total} = W_{1} + W_{2}\] | Calculate total work done by summing areas under the \( F \) vs. \( x \) graph. |
| 2 | \[W_{1} = F_{1} \times \Delta x_{1} = 4 \times 12 = 48 \, \text{J}\] | The work done on the first section (rectangle 0 to 12 m). |
| 3 | \[W_{2} = \frac{1}{2} \cdot 4 \cdot 8 = 16 \, \text{J}\] | The work done on the second section (triangular area from 12 m to 20 m). |
| 4 | \[W_{total} = 48 + 16 = 64 \, \text{J}\] | Total work done. |
| 5 | \[\text{K.E.} = \frac{1}{2}m v_x^2\] | Relate total work done to kinetic energy gain. |
| 6 | \[64 = \frac{1}{2} \cdot 0.20 \cdot v_x^2\] | Substitute \( m = 0.20 \, \text{kg} \). |
| 7 | \[v_x^2 = 640\] | Solve for \( v_x^2 \). |
| 8 | \[v_x = \sqrt{640}\] | Solve for \( v_x \). |
| 9 | \[\boxed{v_x \approx 25.3 \, \text{m/s}}\] | Calculate the final speed at \( x = 20 \text{ m} \). |
Just ask: "Help me solve this problem."
Which of the following graphs represent an object at rest?
A block hangs from the ceiling by a massless rope. A \( 3.0 \, \text{kg} \) block is attached to the first block and hangs below it on another piece of massless rope. The tension in the top rope is \( 63.0 \, \text{N} \).
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 \( \mu \).
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving 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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