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

1 | PE_{\text{initial}} = 0 | The object is initially on the ground, so its initial potential energy is zero (since potential energy is relative to height above a reference point). |

2 | PE_{\text{max}} = mgH | At maximum height H , all the kinetic energy has been converted to potential energy. Potential energy at height H is given by mgH , where m is the mass, g is the acceleration due to gravity, and H is the height. |

3 | PE_{\text{half}} = mg \frac{H}{2} | At half the maximum height \frac{H}{2} , the potential energy is mg \frac{H}{2} . |

4 | KE_{\text{initial}} = mgH | Using the principle of conservation of mechanical energy, the initial kinetic energy must equal the potential energy at maximum height, since there is no kinetic energy at that point (the object is momentarily stationary). |

5 | KE_{\text{half}} = KE_{\text{initial}} – PE_{\text{half}} | To find the kinetic energy at height \frac{H}{2} , subtract the potential energy at that height from the total initial mechanical energy. |

6 | KE_{\text{half}} = mgH – mg \frac{H}{2} = mg \frac{H}{2} | Simplifying, the kinetic energy at \frac{H}{2} is mg \frac{H}{2} , or half the initial kinetic energy. |

7 | ME_{\text{initial}} = ME_{\text{half}} = KE_{\text{initial}} | Total mechanical energy is conserved (since air resistance is negligible). The mechanical energy at any point during the motion is the sum of potential and kinetic energies and is equal to the initial total mechanical energy. |

8 | (a) False, (b) False, (c) False, (d) True, (e) False | Evaluating each statement: (a) Potential energy at \frac{H}{2} is not half of the initial (it’s \frac{1}{2} mgH ). (b) Speed is not directly proportional in this manner. (c) Total mechanical energy is constant, not halved. (d) Kinetic energy at \frac{H}{2} is indeed half of the initial kinetic energy, as shown in the calculations. (e) Since (d) is true, (e) is false. |

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- Statistics

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

MCQ

A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?

- Energy

Beginner

Mathematical

GQ

You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of 1.25 m in a time of 0.84 s. The readout on the display indicates that the average power you are producing is 76 W. What is the magnitude of the force that you exert on the handle?

- Energy

Intermediate

Mathematical

MCQ

At time t = 0, a cart is at x = 10 m and has a velocity of 3 m/s in the –*x* direction. The cart has a constant acceleration in the +*x* direction with magnitude 3 m/s^{2} < a < 6 m/s^{2}?. Which of the following gives the possible range of the position of the cart at t = 1 s?

- 1D Kinematics

Intermediate

Conceptual

MCQ

Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, 4, start rolling down the hill. A few seconds later, Alex’ partner, Bob starts the second ball, B, down the hill by giving it a push. Ball B rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:

- 1D Kinematics

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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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