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Part A – Velocity of block
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \frac{1}{2} m v_x^2 = m g (3.0) \) | Apply conservation of energy. The gravitational potential energy lost by the block when dropped from \(3.0\) m is converted entirely into kinetic energy just before impact. |
| 2 | \( \frac{1}{2} v_x^2 = g (3.0) \) | Cancel the mass \(m\) (since it appears on both sides) to isolate the velocity term. |
| 3 | \( v_x^2 = 2 g (3.0) \) | Multiply both sides by 2 to solve for \(v_x^2\). |
| 4 | \( v_x = \sqrt{2 \times 9.8 \times 3.0} \) | Substitute \(g = 9.8\ \text{m/s}^2\) into the equation. |
| 5 | \( v_x = \sqrt{58.8} \approx 7.67\ \text{m/s} \) | Calculate the square root to determine the final speed just before the spring is struck. |
| 6 | \( \boxed{v_x \approx 7.67\ \text{m/s}} \) | This is the final answer for part (a). |
Part B – Spring Energy
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \Delta h = 3.0\,\text{m} + 0.50\,\text{m} = 3.5\,\text{m} \) | The block drops 3.0 m before striking the spring and an additional \(0.50\,\text{m}\) while compressing the spring, so the total vertical drop is \(3.5\,\text{m}\). |
| 2 | \( U_{\text{spring}} = m g \Delta h \) | The total elastic potential energy stored in the spring equals the gravitational potential energy lost by the block over the total drop \(\Delta h\). |
| 3 | \( U_{\text{spring}} = 240 \times 9.8 \times 3.5 \) | Substitute \(m = 240\,\text{kg}\), \(g = 9.8\,\text{m/s}^2\), and \(\Delta h = 3.5\,\text{m}\) into the energy formula. |
| 4 | \( U_{\text{spring}} \approx 8232\,\text{J} \) | Performing the multiplication, \(240 \times 9.8 = 2352\) and \(2352 \times 3.5 \approx 8232\), yielding the stored elastic energy. |
| 5 | \( \boxed{U_{\text{spring}} \approx 8232\,\text{J}} \) | This is the final answer for part (b): the total elastic potential energy stored in the spring. |
Part C – Spring Constant
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \frac{1}{2} k (\Delta x)^2 = U_{\text{spring}} \) | The elastic potential energy stored in a compressed spring is given by \(\frac{1}{2} k (\Delta x)^2\), where \(\Delta x\) is the compression distance. |
| 2 | \( \frac{1}{2} k (0.50)^2 = 8232 \) | Substitute \(\Delta x = 0.50\,\text{m}\) and \(U_{\text{spring}} = 8232\,\text{J}\) into the formula. |
| 3 | \( \frac{1}{2} k \times 0.25 = 8232 \) | Calculate \((0.50)^2 = 0.25\) and rewrite the equation. |
| 4 | \( 0.125 k = 8232 \) | Simplify \(\frac{1}{2} \times 0.25\) to \(0.125\). |
| 5 | \( k = \frac{8232}{0.125} = 8232 \times 8 \) | Solve for \(k\) by dividing both sides by \(0.125\); note that \(\frac{1}{0.125} = 8\). |
| 6 | \( k \approx 65856\,\text{N/m} \) | Multiply to find the spring constant. |
| 7 | \( \boxed{k \approx 6.59 \times 10^4\,\text{N/m}} \) | This is the final answer for part (c), rounded to three significant figures. |
Just ask: "Help me solve this problem."
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A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
Two masses, \( m_1 \) and \( m_2 \), are connected by a cord and arranged as shown in the diagram, with \( m_1 \) sliding along a frictionless surface and \( m_2 \) hanging from a light, frictionless pulley. What would be the mass of the falling mass, \( m_2 \), if both the sliding mass, \( m_1 \), and the tension, \( T \), in the cord were known?
A piece of metal of weight\(W\) is suspended by two identical strings. Each string passes through a pulley and is attached to a block of mass \(m\) . The system is in equilibrium.What must be true for \(m\) such that the two strings attached to the piece of metal are almost horizontal.
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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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