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# Part (a): Time Elapsed from Leaving the Table to Hitting the Floor
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] t = \sqrt{\frac{2h}{g}} [/katex] | Since the motion in the y-direction is a free fall, we use the kinematic equation [katex] y = \frac{1}{2}gt^2 [/katex] for the vertical motion. Solving for [katex] t [/katex] gives the time it takes to fall a distance [katex] h [/katex]. |
# Part (b): Horizontal Component of the Velocity of the Block Just Before It Hits the Floor
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v_x = \frac{D}{t} [/katex] | Using the result from part (a) and substituting [katex] t = \sqrt{\frac{2h}{g}} [/katex], we get [katex] v_x = \frac{D}{\sqrt{\frac{2h}{g}}} = \sqrt{\frac{D^2g}{2h}} [/katex]. This is the velocity necessary to cover horizontal distance [katex] D [/katex] in time [katex] t [/katex]. |
| 2 | [katex] v_x = \frac{D}{\sqrt{\frac{2h}{g}}} [/katex] | Replace [katex] t [/katex] with the equation found in part a. |
# Part (c): Work Done on the Block by the Spring
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] El = KE [/katex] | The work done by the spring (elastic energy) is transforms into kinetic energy. Since we found the velocity in the previous part we can solve for the kinetic energy. |
| 2 | [katex] El = KE = \frac{1}{2}mv^2[/katex] | Formula for kinetic energy |
| 3 | [katex] KE = \frac{1}{2} m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2 [/katex] | Substitute in velocity found from previous step. |
| 4 | [katex] KE = \frac{mgD^2}{4h}[/katex] | Simplify equation. Note that the Kinetic energy is the work done by the spring. |
# Part (d): Spring Constant
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \frac{1}{2}kx^2 = \frac{1}{2}mv^2 [/katex] | The spring energy ([katex]EL[/katex]) is equal to the kinetic energy as mentioned in part c. Hence we can set [katex] EL = KE [/katex] and solve for [katex]k[/katex]. |
| 2 | [katex] k = \frac{mv^2}{x^2}[/katex] | Solve for k |
| 3 | [katex] k = \frac{m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2}{x^2} [/katex] | Substitute in the [katex] v [/katex], to get the final equation in terms of [katex] M, x, D, h, [/katex] |
| 4 | [katex] k = \frac{mD^2 g}{2hx^2} [/katex] | Simplify |
These steps address each part of the query based on principles of mechanics, conservation of energy, and kinematic equations.
Just ask: "Help me solve this problem."
A 0.2 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.4 J. The kinetic energy as a function of position presented by the graph.
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A bullet moving with an initial speed of [katex] v_o [/katex] strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height [katex] h [/katex]. Which of the following statements is true of the collision.
An object at rest suddenly explodes into two fragments (m1 and m2) by an explosion. Fragment m1 acquires 3 times the kinetic energy of the other. What is the ratio of m1 to m2?
An airplane with a speed of \( 97.5 \, \text{m/s} \) is climbing upward at an angle of \( 50.0^\circ \) with respect to the horizontal. When the plane’s altitude is \( 732 \, \text{m} \), the pilot releases a package.
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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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