| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_B = \rho_{\text{water}} g V_{\text{sub}}\] | The buoyant force on the block equals the weight of the displaced water. Here \(F_B\) is the buoyant force, \(\rho_{\text{water}}\) is the density of water, and \(V_{\text{sub}}\) is the submerged volume of the block. |
| 2 | \[mg = \rho_{\text{water}} g \left(\frac{V}{2}\right)\] | Initially, the block floats with half its volume submerged because half is above water. Since it floats in equilibrium, the buoyant force equals the block’s weight \(mg\). |
| 3 | \[mg = \frac{1}{2}\rho_{\text{water}} g V\] | This shows the weight of the wooden block is equal to the weight of water displaced by half the block’s volume. |
| 4 | \[\rho_{\text{water}} g V = 2mg\] | Multiplying both sides by \(2\) gives the buoyant force when the entire block volume \(V\) is submerged. |
| 5 | \[F_{B,\text{now}} = \rho_{\text{water}} g V\] | After objects are stacked on top, the wooden block is completely submerged, but the objects are not submerged. Therefore, only the full volume of the wooden block displaces water. |
| 6 | \[\boxed{F_{B,\text{now}} = 2mg}\] | The buoyant force acting on the block is now twice the weight of the block alone. Therefore, the correct answer is choice \((c)\). |
| 7 | \[\boxed{\text{Choice }(c)}\] | Choice \((a)\) is incorrect because \(mg\) was the buoyant force only when half the block was submerged. Choice \((b)\) is incorrect because a fully submerged block displaces exactly twice as much water as before, not less than \(2mg\). Choice \((d)\) is incorrect because the buoyant force cannot exceed \(\rho_{\text{water}}gV\) while only the wooden block is submerged. |
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Two blocks of the same size are floating in a container of water. The first block is submerged \( 80\% \) while the second block is submerged by \( 20\% \) beneath the water. Which of the following is a correct statement about the two blocks?
Why do you float higher in salt water than in fresh water?
The \( 70 \) \( \text{kg} \) student in the figure balances a \( 1200 \) \( \text{kg} \) elephant on a hydraulic lift. Assume that it is filled with oil, which is incompressible and has a density \( \rho = 900 \) \( \text{kg/m}^3 \). What is the diameter of the piston the student is standing on? Assume each piston has a cylindrical shape, i.e., a circular cross-sectional area. Note: The two pistons are at the same height. Also, the diameter of the wider piston is given in the figure to be \( 2.0 \) \( \text{m} \).
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of \( 0.5 \) \( \frac{\text{m}}{\text{s}} \) through a \( 2 \) \( \text{cm} \) diameter pipe in the basement under a pressure of \( 3 \) \( \text{atm} \), what will be the flow speed and pressure in a \( 1.3 \) \( \text{cm} \) diameter pipe on the second floor \( 5 \) \( \text{m} \) above?
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
Nancy is using a turkey baster (a kitchen tool with a rubber bulb on one end and a tube on the other) to collect juices from a roasting turkey. When she squeezes and then releases the rubber bulb, it creates suction with a pressure of \( 99{,}800 \) \( \text{Pa} \). This suction causes the turkey juice to rise \( 9 \) \( \text{cm} \) up the tube. Based on this information, what is the density of the turkey juice?
A geologist suspects that her rock specimen is hollow, so she weighs the specimen in both air and water. When completely submerged, the rock weighs twice as much in air as it does in water.
A spherical balloon of mass \( 226 \) \( \text{kg} \) is filled with helium gas until its volume is \( 325 \) \( \text{m}^3 \). Assume the density of air is \( 1.29 \) \( \text{kg/m}^3 \) and the density of helium is \( 0.179 \) \( \text{kg/m}^3 \).
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 \( 1000 \) \( \text{kg/m}^3 \).
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?
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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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