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. |
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An object is suspended from a spring scale first in air, then in water, as shown in the figure above. The spring scale reading in air is \( 17.8 \) \( \text{N} \), and the spring scale reading when the object is completely submerged in water is \( 16.2 \) \( \text{N} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.
A sphere of mass \(0.5\) \(\text{kg}\) is dropped into a column of oil. At the instant the sphere becomes completely submerged in the oil, the sphere is moving downward at \(8\) \(\text{m/s}\), the buoyancy force on the sphere is \(4.0\) \(\text{N}\), and the fluid frictional force is \(4.0\) \(\text{N}\). Which of the following describes the motion of the sphere at this instant?
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
A box of mass \(m\) is initially at rest at the top of a ramp that is at an angle \(\theta\) with the horizontal. The block is at a height \(h\) and length \(L\) from the bottom of the ramp. The coefficient of kinetic friction between the block and the ramp is \(\mu\). What is the kinetic energy of the box at the bottom of the ramp?
A uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
A boulder is raised above the ground so that its potential energy is \(550 \, \text{J}\). Then it is dropped. Assuming \(92 \, \text{J}\) of energy was lost to air resistance, what is the kinetic energy of the boulder just before it hits the ground?
A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).
| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
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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] |
Metric Prefixes
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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