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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."
When the brakes of an automobile are applied, the road exerts the greatest retarding force
For linear motion the term “inertia” refers to the same physical concept of
A skier with a mass of 58 kg glides up a snowy incline that forms an angle of 28 degrees with the horizontal. The skier initially moves at a speed of 7.2 m/s. After traveling a distance of 2.3 meters up the slope, the skier’s speed reduces to 3.8 m/s.
A skydiver reaches a terminal velocity of 55.0 m/s. At terminal velocity, the skydiver no longer accelerates. The mass of the skydiver and her equipment is 87.0 kg. What is the force of friction acting on her?
A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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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