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(a) Calculate the maximum speed of the car at point \( P \).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ PE_A = mgh_A \] | Calculate the potential energy at point \( A \) where \( h_A = 90 \, \text{m} \). |
| 2 | \[ KE_P = \frac{1}{2}mv_{\text{max}}^2 \] | Calculate the kinetic energy at point \( P \). The speed is at maximum here, so \(\Delta h = 0\). |
| 3 | \[ PE_A = KE_P \] | Use the conservation of energy: all potential energy at \( A \) converts into kinetic energy at \( P \). |
| 4 | \[ mgh_A = \frac{1}{2}mv_{\text{max}}^2 \] | Set the potential energy equal to the kinetic energy. The mass \( m \) cancels out. |
| 5 | \[ v_{\text{max}} = \sqrt{2gh_A} \] | Solve for \( v_{\text{max}} \) by rearranging and simplifying the equation. |
| 6 | \[ v_{\text{max}} = \sqrt{2 \times 9.8 \times 90} \] | Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h_A = 90 \, \text{m} \) into the equation. |
| 7 | \[\boxed{v_{\text{max}} = 42 \, \text{m/s}}\] | Calculate the final value of the maximum velocity \( v_{\text{max}} \). |
(b) Calculate the speed \( v_B \) at the top of the loop.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ PE_B = mgh_B \] | Calculate the potential energy at point \( B \) where \( h_B = 50 \, \text{m} \). |
| 2 | \[ KE_B = \frac{1}{2}mv_B^2 \] | Calculate the kinetic energy at point \( B \). |
| 3 | \[ PE_A = PE_B + KE_B \] | Apply conservation of energy between points \( A \) and \( B \). |
| 4 | \[ mgh_A = mgh_B + \frac{1}{2}mv_B^2 \] | Set the potential and kinetic energy sum at \( B \) equal to the potential energy at \( A \). Mass \( m \) cancels out. |
| 5 | \[ gh_A = gh_B + \frac{1}{2}v_B^2 \] | Mass cancels out; simplify the equation for \( v_B \). |
| 6 | \[ v_B^2 = 2g(h_A – h_B) \] | Rearrange to solve for the speed \( v_B \). |
| 7 | \[ v_B = \sqrt{2 \times 9.8 \times (90 – 50)} \] | Substitute values \( g = 9.8 \, \text{m/s}^2 \), \( h_A = 90 \, \text{m} \), and \( h_B = 50 \, \text{m} \) into the equation. |
| 8 | \[\boxed{v_B = 28 \, \text{m/s}}\] | Calculate the speed \( v_B \) at the top of the loop. |
Forces at Point B:
– Normal Force (\( F_N \)): Points towards the center of the loop.
– Gravitational Force (\( F_g \)): Points downwards (also towards the center when car is at the top of the loop).
(c) Calculate the magnitude of all forces and explain loop modification.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ F_N + F_g = \frac{mv_B^2}{r} \] | The centripetal force required at point \( B \). Both forces point towards the center. |
| 2 | \[ F_g = mg \] | Substitute gravitational force where \( m = 700 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). |
| 3 | \[ F_N + mg = \frac{mv_B^2}{r} \] | Substitute the gravitational force into the centripetal force equation. |
| 4 | \[ F_N = \frac{mv_B^2}{r} – mg \] | Rearrange to solve for the normal force \( F_N \). |
| 5 | \[ F_N = \frac{700 \times 28^2}{20} – 700 \times 9.8 \] | Substitute known values: \( m = 700 \, \text{kg} \), \( v_B = 28 \, \text{m/s} \), \( r = 20 \, \text{m} \). |
| 6 | \[\boxed{F_N = 20580 \, \text{N}}\] | Calculate the normal force at the top of the loop. |
Loop Modification:
If the loop’s shape or radius is modified, the drop height should be adjusted accordingly to ensure energy conservation and keep \( v_B \) unchanged. This is supported by the equation derived in part B: \(v_B = \sqrt{2g (h_A – h_B)}\) which shows that \( v_B \) depends only on the height difference of the rollercoaster.
Just ask: "Help me solve this problem."
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a wet patch in the road decreases the maximum static frictional force to one-third its dry road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
A car traveling to the right with a speed \( v \) brakes to a stop in a distance \( d \). What is the work done on the car by the frictional force \( F \)? (Assume that the frictional force is constant)
How does the speed v1 of a block m reaching the bottom of slide 1 compare with v2, the speed of a block 2m reaching the end of slide 2? The blocks are released from the same height.
The Moon does not crash into the Earth because:
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
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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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