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# (a) Direction of the resultant force acting on the marble at point C
The resultant force acting on the marble at point C is directed towards the center of the loop. This force is mainly comprised of the gravitational force pulling downwards and the normal force exerted by the track which also points towards the center of the loop during the circular motion.
# (b) Names of all the forces acting on the marble at point C
| Force | Description |
|---|---|
| Gravitational Force | The force due to gravity acting downwards towards the center of the earth. |
| Normal Force | The force exerted by the surface of the loop on the marble directed radially inward, toward the center of the loop. |
# (c) Deduce the speed of the marble at point C. The working below uses two seperate conservation of energy equations. However, it can also be done in a single equation such that the postential energy at A transfroms into the potential energy at C and the kinetic energy at C. This is written as [katex] mgh_A = mgh_C + \frac{1}{2}mv^2 [/katex].
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] h_A = 0.8 \, \text{m} [/katex] | Initial height from which the marble is released. |
| 2 | [katex] v_A = 0 \, \text{m/s} [/katex] | Initial velocity (marble is released from rest). |
| 3 | [katex] v_B = \sqrt{2gh_A} [/katex] | Re-arrange and solve for velocity at the bottom of the incline, using conservation of mechanical energy, where [katex] mgh_A = \frac{1}{2}m{v^2}_B [/katex]. |
| 4 | [katex] v_B = \sqrt{2 \times 9.8 \times 0.8} [/katex] | Calculating [katex] v_B [/katex]. |
| 5 | [katex] v_B \approx 3.97 \, \text{m/s} [/katex] | Approximate calculation of velocity at point B. |
| 6 | [katex] h_C = 0.35 \, \text{m} [/katex] | The maximum height attained by the marble is at point C (top of loop). |
| 7 | [katex] v_C^2 = v_B^2 – 2gh_C [/katex] | Using conservation of mechanical energy between points B and C. |
| 8 | [katex] v_C^2 = 3.97^2 – 2 \times 9.8 \times 0.35 [/katex] | Calculating [katex] v_C [/katex] from [katex] v_B [/katex] and change in gravitational potential energy. |
| 9 | [katex] v_C \approx 3.0 \, \text{m/s} [/katex] | Approximate calculation of velocity at point C. |
# (d) Effect if the release height of the marble were to double
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( mgh_A = mgh_C + \frac{1}{2}mv^2 \) | The conservation of energy between points A and C, as used in part B. |
| 2 | \( gh_A = gh_C + \frac{1}{2}v^2 \) | Simplified formula, canceling out \( m \) on both sides. |
| 3 | \( g\Delta h = \frac{1}{2}v^2 \) | Replace \( gh_A – gh_C \) with \( g\Delta h \), which represents the change in height. |
| 4 | \( \frac{\Delta h}{v^2} = \frac{1}{2g} \) | Isolate \( \Delta h \) and \( v^2 \) to see their proportional relationship. |
| 5 | Proportional analysis | In the initial setup: \( \Delta h = 0.8 – 0.35 = 0.45 \, \text{m} \). Doubling the release height, \( \Delta h \) becomes \( 1.6 – 0.35 = 1.25 \, \text{m} \). The ratio of \( \Delta h \) is: \[ \frac{1.25}{0.45} \approx 2.78 \] Since \( v^2 \propto \Delta h \), the velocity increases by \( \sqrt{2.78} \approx 1.67 \). |
| 6 | Conclusion | The original velocity was \( 3 \, \text{m/s} \). With a velocity ratio of \( 1.67 \), the final velocity becomes: \[ 3 \times 1.67 \approx 5 \, \text{m/s} \] |
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A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude \( A \) on a horizontal frictionless surface. Determine the following in terms of only the variable \( A \).
| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
A race car traveling at a constant speed of \( 50 \) \( \text{m/s} \) drives around a circular track that is \( 500 \) \( \text{m} \) in diameter. What is the magnitude of the acceleration of the car?
Consider an object on a rotating disk at a distance \( r \) from its center, held in place on the disk by static friction. Which of the following statements is not true concerning this object?
The box in the diagram is sliding to the right across a horizontal table, under the influence of the forces shown. Which force(s) is doing negative work on the box?
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 mechanic pushes a [katex]2500 \, \text{kg}[/katex] car from rest to a final speed [katex]v[/katex] by doing [katex]5.0 \times 10^3 \, \text{J}[/katex] of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
A \(5.0 \, \text{g}\) coin is placed \(15 \, \text{cm}\) from the center of a turntable. The coin has coefficients of static and kinetic friction of \(\mu_s = 0.80\) and \(\mu_k = 0.50\). The turntable slowly speeds up to \(60 \, \text{rpm}\). Does the coin slide off the turntable?
A ball is thrown straight up. At what point does the ball have the most energy?
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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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