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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] KE_{\text{peak}} = \frac{1}{2} m v^2 [/katex] | Calculate the kinetic energy at the peak height. Kinetic energy (KE) is given by the equation [katex]\frac{1}{2} m v^2[/katex], where [katex]m[/katex] is mass (2 kg) and [katex]v[/katex] is velocity (150 m/s). |
| 2 | [katex] PE_{\text{peak}} = mgh [/katex] | Calculate the potential energy at the peak height. Potential energy (PE) is given by [katex]mgh[/katex] where [katex]m[/katex] is mass (2 kg), [katex]g[/katex] is the acceleration due to gravity (approximately [katex]9.8 \, \text{m/s}^2[/katex]) and [katex]h[/katex] is the height (90 m). |
| 3 | [katex] W_{\text{thrust}} = F_{\text{thrust}} \times h [/katex] | Calculate the work done by the thrust force. Work ([katex]W[/katex]) is force ([katex]F[/katex]) times the displacement ([katex]h[/katex]). Here, [katex]F_{\text{thrust}}[/katex] is 275 N. |
| 4 | [katex] KE_{\text{peak}} = \frac{1}{2} \times 2 \times 150^2 = 22500 \, \text{J} [/katex] | Substitute the values to find [katex] KE_{\text{peak}} [/katex]. |
| 5 | [katex] PE_{\text{peak}} = 2 \times 9.8 \times 90 = 1764 \, \text{J} [/katex] | Substitute the values to find [katex] PE_{\text{peak}} [/katex]. |
| 6 | [katex] W_{\text{thrust}} = 275 \times 90 = 24750 \, \text{J} [/katex] | Substitute the values to find [katex] W_{\text{thrust}} [/katex]. |
| 7 | [katex] W_{\text{thrust}} = KE_{\text{peak}} + PE_{\text{peak}} + W_{AR} [/katex] | Use the conservation of energy. In this case the initial work done by the rocket thrust transforms into kinetic energy, potential energy, and work done by air resistance [katex] W_{AR} [/katex]. |
| 8 | [katex] W_{AR} = 24750 – 22500 – 1764 [/katex] | Solving for the work done by air resistance gives us [katex] W_{AR} = 486 \, \text{J} [/katex]. |
| 9 | [katex] W_{AR} = Fh[/katex] | Divide the work done by air resistance by the height to find the average air resistance force. |
| 10 | [katex] W_{AR} = \frac{486}{90} \approx 5.4 \, \text{N} [/katex] | [katex] F_{\text{air}} \approx \boxed{5.4} \, \text{N} [/katex] is the average air resistance force acting against the rocket during ascent. |
ALTERNATE EXPLANATION – Using Forces
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{net}} = ma [/katex] | Newton’s second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. |
| 2 | [katex] F_{\text{net}} = F_{\text{thrust}} – F_{\text{gravity}} – F_{\text{drag}} [/katex] | The net force acting on the rocket includes the thrust force, the gravitational force, and the drag force (air resistance), where the thrust and drag act upwards, and gravity acts downwards. |
| 3 | [katex] F_{\text{gravity}} = mg [/katex] | The gravitational force is calculated as the product of the mass and the acceleration due to gravity [katex] g \approx 9.81 \, \text{m/s}^2 [/katex]. |
| 4 | [katex] F_{\text{gravity}} = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} [/katex] | Substitute the values of the mass and gravitational acceleration to find the gravitational force. |
| 5 | [katex] a = \frac{v^2 – v_0^2}{2h} [/katex] | The kinematic equation relates the rocket’s change in velocity to acceleration and distance. Here, [katex] v_0 = 0 \, \text{m/s} [/katex] (initial velocity), [katex] v = 150 \, \text{m/s} [/katex] (final velocity at the peak), and [katex] h = 90 \, \text{m} [/katex] (height). |
| 6 | [katex] a = \frac{(150 \, \text{m/s})^2}{2 \times 90 \, \text{m}} [/katex] | Substituting the values into the kinematic equation to calculate acceleration. |
| 7 | [katex] a = 125 \, \text{m/s}^2 [/katex] | Calculation of the acceleration using the given values. |
| 8 | [katex] F_{\text{net}} = 2 \, \text{kg} \times 125 \, \text{m/s}^2 = 250 \, \text{N} [/katex] | Calculate the net force acting on the rocket using Newton’s second law with the rocket’s mass and the acceleration. |
| 9 | [katex] 250 \, \text{N} = 275 \, \text{N} – 19.62 \, \text{N} – F_{\text{drag}} [/katex] | Solve for the drag force using the equation in step 2. |
| 10 | [katex] F_{\text{drag}} = 275 \, \text{N} – 19.62 \, \text{N} – 250 \, \text{N} = 5.38 \, \text{N} [/katex] | Substitute the values of the thrust, gravitational, and net forces to find the drag force. |
| 11 | [katex] F_{\text{drag}} \approx 5.38 \, \text{N} [/katex] | This is the average air resistance force acting on the rocket during its ascent. |
Just ask: "Help me solve this problem."
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s2. How far up does it move?
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
A bullet moving with an initial speed of [katex] v_o [/katex] strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height [katex] h [/katex]. Which of the following statements is true of the collision.
[katex] F_{\text{air}} \approx 5.4 \, \text{N} [/katex]
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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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