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

1 | F_{\text{parallel}} = mg\sin(\theta) | Force parallel to the incline, where m is mass, g is acceleration due to gravity, and \theta is the incline angle. |

2 | F_{\text{normal}} = mg\cos(\theta) | Normal force, perpendicular to the incline. |

3 | F_{\text{friction}} = \mu_k F_{\text{normal}} | Frictional force, where \mu_k is the coefficient of kinetic friction. |

4 | F_{\text{friction}} = 0.250 \times mg\cos(25^\circ) | Substitute values for \mu_k and F_{\text{normal}}. |

5 | F_{\text{pull}} = F_{\text{parallel}} + F_{\text{friction}} | Total force to pull the crate includes parallel force and frictional force. |

6 | F_{\text{pull}} = mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) | Combine parallel and frictional forces. |

7 | W_{\text{output}} = F_{\text{parallel}} \times d | Work output, where F_{\text{parallel}} is the force parallel to the incline and d is the distance. |

8 | W_{\text{output}} = mg\sin(25^\circ) \times 2.5 | Work done against gravity. |

9 | W_{\text{input}} = F_{\text{pull}} \times d | Work input, force to pull the crate. |

10 | W_{\text{input}} = \left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5 | Work done to pull the crate including overcoming friction. |

11 | \text{Efficiency} = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100% | Efficiency formula. |

12 | \text{Efficiency} = \frac{mg\sin(25^\circ) \times 2.5}{\left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5} \times 100% | Substitute W_{\text{output}} and W_{\text{input}}. |

13 | \text{Efficiency} = \frac{\sin(25^\circ)}{\sin(25^\circ) + 0.250 \times \cos(25^\circ)} \times 100% | Simplify by canceling mg and d. |

14 | \text{Efficiency} = 65.10\% | Calculated efficiency. |

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- Statistics

Advanced

Mathematical

GQ

A uniform solid cylinder of mass M and radius R is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force F , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \theta , what is the kinetic energy of the cylinder in terms of F and \theta ?

- Energy, Rotational Kinematics, Rotational Motion

Beginner

Conceptual

MCQ

A ball is thrown straight up. At what point does the ball have the most energy?

- Energy

Intermediate

Mathematical

FRQ

A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.

- Energy, Simple Harmonic Motion, Springs

Advanced

Conceptual

MCQ

In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?

- Energy

Intermediate

Mathematical

FRQ

A small block moving with a constant speed v collides inelastically with a block M attached to one end of a spring k. The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.

- Energy, Simple Harmonic Motion, Springs

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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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