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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Jill does twice as much work as Jack does and in half the time. Jill’s power output is
A bullet (mass: \(0.05 \, \text{kg}\)) is fired horizontally (\(v = 200 \, \text{m/s}\)) at a block (mass: \(1.3 \, \text{kg}\)) initially at rest on a frictionless surface. The block is attached to a spring (\(k = 2500 \, \text{N/m}\)). The bullet becomes embedded. Calculate:
A skier with a mass of \(58 \, \text{kg}\) glides up a snowy incline that forms an angle of \(28^\circ\) with the horizontal. The skier initially moves at a speed of \(7.2 \, \text{m/s}\). After traveling a distance of \(2.3 \, \text{m}\) up the slope, the skier’s speed reduces to \(3.8 \, \text{m/s}\).
Two masses \(m_1\) and \(4m_1\) are on an incline. Both surfaces have the same coefficient of kinetic friction. Both objects start from rest at the same height. Which mass has the largest speed at the bottom?
A bullet moving with an initial speed of \( v_o \) strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height \( h \). Which of the following statements is true of the collision.
A sample of an unknown material appears to weigh \( 285 \) \( \text{N} \) in air and \( 195 \) \( \text{N} \) when immersed in alcohol of specific gravity \( 0.700 \).
Johnny the auto mechanic is raising a \( 1200 \) \( \text{kg} \) car on her hydraulic lift so that she can work underneath. If the area of the input piston is \( 12 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
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?
Wanda watches the fish in her fish tank and notices that the angelfish like to feed at the water’s surface, while the catfish feed \( 0.300 \) \( \text{m} \) below at the bottom of the tank. If the average density of the water in the tank is \( 1000\) \( \text{kg/m}^3 \), what is the absolute pressure on the catfish?
A typical \( 68 \text{-kg} \) person generates a steady mechanical power output of \( 120 \text{ W} \) at the pedals of a bicycle. Approximately how many Calories are “burned” (total metabolic energy expended) when the person rides a bicycle for \( 15 \text{ minutes} \)? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy. Note (\( 1 \text{ Cal} = 4186 \text{ J} \)).
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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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