| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (a): Determining the Spring Constant \(k\) | ||
| 1 | \[\Delta x_{\text{air}} = L_{\text{stretched}} – L_{\text{unstretched}}\] | Measure the length of the spring without the object and then with the object at rest in air. Their difference is the displacement \(\Delta x_{\text{air}}\). |
| 2 | \[mg = k\,\Delta x_{\text{air}}\] | At equilibrium in air, the downward gravitational force \(mg\) is balanced by the spring force \(k\,\Delta x_{\text{air}}\). |
| 3 | \[k = \frac{mg}{\Delta x_{\text{air}}}\] | Solve for \(k\) by algebraically rearranging the force equilibrium equation. |
| Alternative Method: One can also determine \(k\) by setting the mass into oscillation and using \(T = 2\pi\sqrt{\frac{m}{k}}\) to solve for \(k = \frac{4\pi^2m}{T^2}\), but the displacement method is straightforward using a metric ruler. | ||
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (b): Changes When the Object is Immersed | ||
| 1 | \[mg – F_{b} = k\,\Delta x_{\text{fluid}}\] | When the object is immersed in the fluid, it experiences an upward buoyant force \(F_{b}\). Therefore, the spring now only needs to balance the net force \(mg – F_{b}\), resulting in a smaller displacement \(\Delta x_{\text{fluid}}\) compared to \(\Delta x_{\text{air}}\). |
| 2 | \(\Delta x_{\text{fluid}} < \Delta x_{\text{air}}\) | The observed change is a decrease in the spring extension because the fluid’s buoyant force partially offsets the weight of the object. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (c): Experimental Determination of Fluid Density \(\rho\) | ||
| 1 | Measure \(\Delta x_{\text{air}}\) and \(\Delta x_{\text{fluid}}\). | Using the metric ruler, record the spring displacement when the object is in air and when it is immersed in the fluid. |
| 2 | Determine \(k = \frac{mg}{\Delta x_{\text{air}}}\) from Part (a). | This value of \(k\) is required for the next step of finding the buoyant force. |
| 3 | Measure the object’s mass \(m\) and use its known density \(D\) to find its volume \(V\) via \(V = \frac{m}{D}\). | By definition, density is mass per unit volume. Since \(D \gg \rho\), the object is practically incompressible and its volume can be calculated accurately. |
| 4 | Relate the buoyant force and the displaced fluid: \(F_{b} = \rho\,V\,g\). | According to Archimedes’ principle, the buoyant force equals the weight of the displaced fluid. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| Part (d): Using Measurements to Calculate \(\rho\) | ||
| 1 | \(mg = k\,\Delta x_{\text{air}}\) | At equilibrium in air, the gravitational force is balanced by the spring force. |
| 2 | \(mg – k\,\Delta x_{\text{fluid}} = \rho\,V\,g\) | In the fluid, the buoyant force \(\rho\,V\,g\) reduces the effective force the spring must support. |
| 3 | Substitute \(k = \frac{mg}{\Delta x_{\text{air}}}\): \(mg – \frac{mg}{\Delta x_{\text{air}}}\,\Delta x_{\text{fluid}} = \rho\,V\,g\) | This substitution expresses the equation in terms of measurable quantities \(\Delta x_{\text{air}}\) and \(\Delta x_{\text{fluid}}\). |
| 4 | Divide by \(g\): \(m\Bigl(1- \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr) = \rho\,V\) | Simplify the equation by eliminating the gravitational acceleration \(g\), which appears on both sides. |
| 5 | Solve for \(\rho\): \(\displaystyle \rho = \frac{m \Bigl(1- \dfrac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)}{V}\) | Isolate \(\rho\) to relate it directly to the measurements and the known mass and volume of the object. |
| 6 | Substitute \(V = \frac{m}{D}\): \(\displaystyle \rho = \frac{m \Bigl(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)}{\frac{m}{D}} = D\Bigl(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\Bigr)\) | Since the object\’s density \(D\) and mass \(m\) give its volume, this substitution yields the final formula for the fluid density \(\rho\) in terms of \(D\) and the measured displacements. |
| 7 | \[\boxed{\rho = D \left(1 – \frac{\Delta x_{\text{fluid}}}{\Delta x_{\text{air}}}\right)}\] | This is the explicit expression to calculate the fluid density based on the measured values. |
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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 \( 0.700 \).
In a town’s water system, pressure gauges in still water at street level read \( 150 \) \( \text{kPa} \). If a pipeline connected to the system breaks and shoots water straight up, how high above the street does the water shoot?
The experimental diving rig is lowered from rest at the ocean’s surface and reaches a maximum depth of \(80\) \(\text{m}\). Initially it accelerates downward at a rate of \(0.10\) \(\text{m/s}^2\) until it reaches a speed of \(2.0\) \(\text{m/s}\), which then remains constant. During the descent, the pressure inside the bell remains constant at \(1\) atmosphere. The top of the bell has a cross-sectional area \(A = 9.0\) \(\text{m}^2\). The density of seawater is \(1025\) \(\text{kg/m}^3\).
Which of the following statements is an expression of the equation of continuity?
Alcohol has a specific gravity of \( 0.79 \). If a barometer consisting of an open-ended tube placed in a dish of alcohol is used at sea level, to what height in the tube will the alcohol rise?
In a carbonated drink dispenser, bubbles flow through a horizontal tube that gradually narrows in diameter. Assuming the change in height is negligible, which of the following best describes how the bubbles behave as they move from the wider section of the tube to the narrower section?
If the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
A Venturi meter is a device used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed \( v_2 \) through a horizontal section of pipe with a cross-sectional area \( A_2 = 542 \) \( \text{cm}^2 \). The gas has a density of \( 1.35 \) \( \text{kg/m}^3 \). The Venturi meter has a cross-sectional area of \( A_1 = 215 \) \( \text{cm}^2 \) and has been substituted for a section of the larger pipe. The pressure difference between the two sections \( P_2 – P_1 = 145 \) \( \text{Pa} \).
A linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
Check the explanation for the complete solution. The following is a condensed version of the solution:
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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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