Step | Derivation/Formula | Reasoning |
---|---|---|

1 | KE_{\text{peak}} = \frac{1}{2} m v^2 | Calculate the kinetic energy at the peak height. Kinetic energy (KE) is given by the equation \frac{1}{2} m v^2, where m is mass (2 kg) and v is velocity (150 m/s). |

2 | PE_{\text{peak}} = mgh | Calculate the potential energy at the peak height. Potential energy (PE) is given by mgh where m is mass (2 kg), g is the acceleration due to gravity (approximately 9.8 \, \text{m/s}^2) and h is the height (90 m). |

3 | W_{\text{thrust}} = F_{\text{thrust}} \times h | Calculate the work done by the thrust force. Work (W) is force (F) times the displacement (h). Here, F_{\text{thrust}} is 275 N. |

4 | KE_{\text{peak}} = \frac{1}{2} \times 2 \times 150^2 = 22500 \, \text{J} | Substitute the values to find KE_{\text{peak}} . |

5 | PE_{\text{peak}} = 2 \times 9.8 \times 90 = 1764 \, \text{J} | Substitute the values to find PE_{\text{peak}} . |

6 | W_{\text{thrust}} = 275 \times 90 = 24750 \, \text{J} | Substitute the values to find W_{\text{thrust}} . |

7 | W_{\text{thrust}} = KE_{\text{peak}} + PE_{\text{peak}} + W_{AR} | 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 W_{AR} . |

8 | W_{AR} = 24750 – 22500 – 1764 | Solving for the work done by air resistance gives us W_{AR} = 486 \, \text{J} . |

9 | W_{AR} = Fh | Divide the work done by air resistance by the height to find the average air resistance force. |

10 | W_{AR} = \frac{486}{90} \approx 5.4 \, \text{N} | F_{\text{air}} \approx \boxed{5.4} \, \text{N} is the average air resistance force acting against the rocket during ascent. |

**ALTERNATE EXPLANATION – Using Forces**

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | F_{\text{net}} = ma | 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 | F_{\text{net}} = F_{\text{thrust}} – F_{\text{gravity}} – F_{\text{drag}} | 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 | F_{\text{gravity}} = mg | The gravitational force is calculated as the product of the mass and the acceleration due to gravity g \approx 9.81 \, \text{m/s}^2 . |

4 | F_{\text{gravity}} = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} | Substitute the values of the mass and gravitational acceleration to find the gravitational force. |

5 | a = \frac{v^2 – v_0^2}{2h} | The kinematic equation relates the rocket’s change in velocity to acceleration and distance. Here, v_0 = 0 \, \text{m/s} (initial velocity), v = 150 \, \text{m/s} (final velocity at the peak), and h = 90 \, \text{m} (height). |

6 | a = \frac{(150 \, \text{m/s})^2}{2 \times 90 \, \text{m}} | Substituting the values into the kinematic equation to calculate acceleration. |

7 | a = 125 \, \text{m/s}^2 | Calculation of the acceleration using the given values. |

8 | F_{\text{net}} = 2 \, \text{kg} \times 125 \, \text{m/s}^2 = 250 \, \text{N} | Calculate the net force acting on the rocket using Newton’s second law with the rocket’s mass and the acceleration. |

9 | 250 \, \text{N} = 275 \, \text{N} – 19.62 \, \text{N} – F_{\text{drag}} | Solve for the drag force using the equation in step 2. |

10 | F_{\text{drag}} = 275 \, \text{N} – 19.62 \, \text{N} – 250 \, \text{N} = 5.38 \, \text{N} | Substitute the values of the thrust, gravitational, and net forces to find the drag force. |

11 | F_{\text{drag}} \approx 5.38 \, \text{N} |
This is the average air resistance force acting on the rocket during its ascent. |

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- Statistics

Intermediate

Mathematical

FRQ

A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s^{2} and is lifted through a distance of 11 m.

- Energy, Linear Forces

Intermediate

Conceptual

MCQ

A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.

Case A: Thrown straight up.

Case B: Thrown straight down.

Case C: Thrown out at an angle of 45° above horizontal.

Case D: Thrown straight out horizontally.

In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?

- Energy

Advanced

Mathematical

GQ

A horizontal force of 110 N is applied to a 12 kg object, moving it 6 m on a horizontal surface where the kinetic friction coefficient is 0.25. The object then slides up a 17° inclined plane. Assuming the 110 N force is no longer acting on the incline, and the coefficient of kinetic friction there is 0.45, calculate the distance the object will slide on the incline.

- Energy

Advanced

Mathematical

FRQ

The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.

- Circular Motion, Energy

Advanced

Mathematical

FRQ

A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.

- Energy, Linear Forces

F_{\text{air}} \approx 5.4 \, \text{N}

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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_1m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu N |

R = \frac{v_i^2 \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |

a_c = \frac{v^2}{r} | PE = mgh |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m v | \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 |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |

M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |

M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kgm/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (Nm)} |

I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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