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To solve the problem using principles of work and energy, we consider the system comprising the two masses. We know that one mass will move up while the other moves down, and we’re given the displacement [katex] d = 0.8 [/katex] meters for each mass.
Assuming the 12 kg mass moves downward and the 9 kg mass moves upward, we need to find the common velocity [katex] v [/katex] after they have moved 0.8 meters, taking into account that energy is conserved in this isolated system (ignoring air resistance and friction). Here’s the step-by-step analysis:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] m_1 = 12 \, \text{kg} [/katex] [katex] m_2 = 9 \, \text{kg} [/katex] |
Define the masses where [katex] m_1 [/katex] is the 12 kg mass and [katex] m_2 [/katex] is the 9 kg mass. |
| 2 | [katex] g = 9.81 \, \text{m/s}^2 [/katex] | Acceleration due to gravity is [katex] g [/katex]. |
| 3 | [katex] \Delta h = 0.8 \, \text{m} [/katex] | Both masses displace by 0.8 m, one moving up and the other down. |
| 4 | [katex] U_{\text{initial}} = 0 \, \text{J} [/katex] | Initial potential energy is set to zero for simplicity as we’re only interested in changes. |
| 5 | [katex] U_{\text{final}} = -m_1 g \Delta h + m_2 g \Delta h [/katex] | Final potential energy obtained by considering the gain in height by [katex] m_2 [/katex] and loss in height by [katex] m_1 [/katex]. |
| 6 | [katex] U_{\text{final}} = -12 \times 9.81 \times 0.8 + 9 \times 9.81 \times 0.8 [/katex] | Substitute the known values to find [katex] U_{\text{final}} [/katex]. |
| 7 | [katex] U_{\text{final}} = -23.544 \, \text{J} [/katex] | Calculate the final potential energy. |
| 8 | [katex] KE_{\text{initial}} = 0 \, \text{J} [/katex] | Initial kinetic energy, assuming they start from rest. |
| 9 | [katex] KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v^2 [/katex] | Kinetic energy for the system of the two moving masses at the final state. |
| 10 | [katex] KE_{\text{final}} = \frac{1}{2} (12+9) v^2 [/katex] | Expression for the final kinetic energy of both masses. |
| 11 | [katex] KE_{\text{final}} = 10.5 v^2 [/katex] | Simplifying the expression for final kinetic energy. |
| 12 | [katex] KE_{\text{final}} = -U_{\text{final}} [/katex] | By conservation of energy, change in potential energy equals change in kinetic energy. |
| 13 | [katex] 10.5 v^2 = 23.544 [/katex] | Substitute computed value of [katex] U_{\text{final}} [/katex]. |
| 14 | [katex] v^2 = \frac{23.544}{10.5} [/katex] | Solve for [katex] v^2 [/katex]. |
| 15 | [katex] v = \sqrt{\frac{23.544}{10.5}} [/katex] | Calculate [katex] v [/katex]. |
| 16 | [katex] v \approx 1.49 \, \text{m/s} [/katex] | Final velocity of the masses after traveling 0.8 m. |
This result indicates the common speed of both masses after they have moved 0.8 meters, based on the conservation of mechanical energy.
Just ask: "Help me solve this problem."
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?
An average adult elephant \( (5000 \, \text{kg}) \) is strapped to a spring, which is then pulled \( 2 \, \text{meters} \) away from its equilibrium position and released. The elephant starts oscillating back and forth with a period of \( 10 \) seconds.
An object with a mass m = 80 g is attached to a spring with a force constant k = 25 N/m. The spring is stretched 52.0 cm and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.
1.49 m/s
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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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