| 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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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 \( 2.0 \) \( \text{m} \).
A \(0.50 \, \text{kg}\) mass is attached to a spring constant \(20 \, \text{N/m}\) along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of \(1.5 \, \text{m/s}\) at the equilibrium position. What is the total energy of the system?
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 \(1025\) \(\text{kg/m}^3\).
A linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
A student designs an experiment to determine the density of an unknown fluid. The student pours the fluid into a graduated cylinder and attaches an object to a force probe. The object has a density greater than the density of the fluid. The student partially submerges the object into the fluid and records both the volume of fluid displaced in the graduated cylinder and the reading on the force probe. The student then submerges the object further and, at each trial, records the new values of displaced volume and force probe reading until the object is fully submerged. The student constructs a graph of force probe reading (vertical axis) as a function of volume of fluid displaced (horizontal axis). Which of the following statements correctly describes how a feature of this graph is related to the density of the fluid?
The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
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?
In the lab, a student is given a glass beaker filled with water with an ice cube of mass \( m \) and volume \( V_c \) floating in it.
The downward force of gravity on the ice cube has magnitude \( F_g \). The student pushes down on the ice cube with a force of magnitude \( F_P \) so that the cube is totally submerged. The water then exerts an upward buoyant force on the ice cube of magnitude \( F_B \). Which of the following is an expression for the magnitude of the acceleration of the ice cube when it is released?
A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
A solid plastic cube with uniform density (side length = \(0.5\) \(\text{m}\)) of mass \(100\) \(\text{kg}\) is placed in a vat of fluid whose density is \(1200\) \(\text{kg/m}^3\). What fraction of the cube’s volume floats above the surface of the fluid?
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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