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Part A – Minimum work required
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = 35\,\text{m} + 50\,\text{m} = 85\,\text{m}\] | The water is pumped from the bottom of the well (35 m deep) to a house 50 m above the well; hence the total vertical displacement is \(85\,\text{m}\). |
| 2 | \[W = \rho\,V\,g\,\Delta x\] | This formula gives the gravitational work (or potential energy gain) required to move a volume \(V\) of water with density \(\rho\) against gravity \(g\) through a vertical displacement \(\Delta x\). |
| 3 | \[W = (1000\,\text{kg/m}^3)(0.35\,\text{m}^3)(9.8\,\text{m/s}^2)(85\,\text{m})\] | Substitute the numerical values: water density \(\rho = 1000\,\text{kg/m}^3\), volume \(V = 0.35\,\text{m}^3\), gravitational acceleration \(g = 9.8\,\text{m/s}^2\), and displacement \(\Delta x = 85\,\text{m}\). |
| 4 | \[W \approx 2.92 \times 10^5\,\text{J}\] | Performing the multiplication gives \(W \approx 291550\,\text{J}\), which is rounded to \(2.92 \times 10^5\,\text{J}\). |
| 5 | \[\boxed{W \approx 2.92 \times 10^5\,\text{J}}\] | This is the minimum work required to pump the water used per day. |
Part B – Minimum power rating
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = 2\,\text{hours} = 7200\,\text{s}\] | Convert the pumping time from hours to seconds for consistency in SI units. |
| 2 | \[P = \frac{W}{t}\] | Power is defined as the work done per unit time. |
| 3 | \[P = \frac{2.92 \times 10^5\,\text{J}}{7200\,\text{s}} \approx 40.5\,\text{W}\] | Substitute the work calculated in part (a) and the conversion for time to find the minimum power rating of the pump. |
| 4 | \[\boxed{P \approx 40.5\,\text{W}}\] | This represents the minimum power needed to pump the water within the specified time. |
Part C – Flow velocity
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[A_{\text{well}} = \frac{\pi (0.03)^2}{4}\] | Calculate the cross-sectional area of the pipe in the well (pipe diameter = 3.0 cm = 0.03 m). |
| 2 | \[Q = A_{\text{well}}\,v_i\] | The volumetric flow rate \(Q\) in the well is the product of the area and the well velocity \(v_i = 0.50\,\text{m/s}\). |
| 3 | \[A_{\text{house}} = \frac{\pi (0.0125)^2}{4}\] | Calculate the cross-sectional area of the pipe at the house (pipe diameter = 1.25 cm = 0.0125 m). |
| 4 | \[A_{\text{well}}\,v_i = A_{\text{house}}\,v_x \quad \Rightarrow \quad v_x = \frac{A_{\text{well}}}{A_{\text{house}}}\,v_i\] | Using mass continuity (volume flow rate is constant), relate the velocity in the well \(v_i\) to the velocity in the house \(v_x\) via their cross-sectional areas. |
| 5 | \[\frac{A_{\text{well}}}{A_{\text{house}}} = \left(\frac{0.03}{0.0125}\right)^2 = (2.4)^2 = 5.76\] | Since both areas involve the factor \(\pi/4\), the ratio simplifies to the square of the ratio of the diameters. |
| 6 | \[v_x = 5.76 \times 0.50 = 2.88\,\text{m/s}\] | Multiply the well velocity by the area ratio to obtain the flow velocity at the house’s faucet. |
| 7 | \[\boxed{v_x \approx 2.88\,\text{m/s}}\] | This is the calculated flow velocity when the faucet in the house is open. |
Part D – Calculating minimum pressure
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Energy per unit volume: } \Delta P = \rho g\,\Delta x + \frac{1}{2}\rho\,(v_x^2 – v_i^2)\] | This expression (derived from Bernoulli’s principle) represents the pressure energy per unit volume needed to overcome both the gravitational potential \(\rho g\,\Delta x\) and to provide the increase in kinetic energy from \(v_i\) (in the well) to \(v_x\) (at the faucet). |
| 2 | \[P_{\text{faucet, min}} = \rho g\,\Delta x + \frac{1}{2}\rho\,v_x^2\] | For minimum pressure at the faucet (assuming negligible initial kinetic energy \(v_i \approx 0\) or that its contribution is already included in the pump pressure), the faucet must supply at least the hydrostatic pressure plus the dynamic pressure necessary for the flow velocity \(v_x\). In practice, if the faucet discharges to the atmosphere, its gauge pressure is 0, so the pump must overcome this total pressure drop. |
| 3 | \[\boxed{P_{\text{min at faucet}} = \rho g\,\Delta x + \frac{1}{2}\rho\,v_x^2}\] | This is how one would calculate the minimum pressure required at the faucet: by summing the pressure needed to lift the water \(\rho g\,\Delta x\) and the pressure needed to accelerate it to \(v_x\) (i.e., \(\frac{1}{2}\rho\,v_x^2\)). |
| 4 | N/A | In summary, to find the minimum pressure at the faucet, you equate the pressure energy per unit volume provided by the pump with the sum of the gravitational and dynamic energy per unit volume required for the water to reach the faucet at velocity \(v_x\). This is essentially an algebraic application of Bernoulli’s principle under ideal (lossless) conditions. |
Just ask: "Help me solve this problem."
Ball A of mass m is dropped from a building of height H. Ball B of mass 1.7m is dropped from a building of height 3.5H. Using energy, what the ratio of vA to vB (final velocity of ball A to final velocity of ball B). Air resistance is negligible.
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
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.
Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance d. Ignore friction and assume that an equal force F is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?
The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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