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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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Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.
A 25.0-kg box is released on a 23.5° incline and accelerates down the incline at 0.35 m/s2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?
A person is making homemade ice cream. She exerts a force of magnitude 23 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.25 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.7 s, what is the average power being expended?
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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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