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| 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. |
Just ask: "Help me solve this problem."
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
A person is standing on a railroad station platform when a high-speed train passes by. The person will tend to be
A \( 1.5 \; \text{kg} \) mass attached to a spring with a force constant of \( 20.0 \; \text{N/m} \) oscillates on a horizontal, frictionless track. At \( t = 0 \), the mass is released from rest at \( x = 10.0 \; \text{cm} \). (That is, the spring is stretched by \( 10.00 \; \text{cm} \).)
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 small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
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] |
General Metric Conversion Chart
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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