| 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 \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
Two points, \( A \) and \( B \), are in a pipe carrying a flowing ideal fluid. Point \( B \) is \( 2.0 \) \( \text{m} \) higher than point \( A \), and the fluid speed at \( B \) is twice the speed at \( A \). If the pressure at \( A \) is \( P_A \), which of the following expressions correctly represents the pressure at \( B \) \( (P_B) \)?
Why do you float higher in salt water than in fresh water?
A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.
A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?
A trash compactor pushes down with a force of \( 500 \) \( \text{N} \) on a \( 3 \) \( \text{cm}^2 \) input piston, causing a force of \( 30,000 \) \( \text{N} \) to crush the trash. What is the area of the output piston that crushes the trash?
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), 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?
An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
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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