**Part a: Calculate the acceleration of the system**

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

1 | F_{\text{net,A}} = m_A \cdot a | Net force on mass A equals mass times acceleration. |

2 | F_{\text{net,B}} = m_B \cdot a | Net force on mass B equals mass times acceleration. |

3 | F_{\text{net,A}} = T – m_A \cdot g | Tension upwards minus weight of A downwards. |

4 | F_{\text{net,B}} = m_B \cdot g – T | Weight of B downwards minus tension upwards. |

5 | m_A \cdot a = T – m_A \cdot g | Substitute step 1 into step 3. |

6 | m_B \cdot a = m_B \cdot g – T | Substitute step 2 into step 4. |

7 | m_A \cdot a + m_B \cdot a = m_B \cdot g – m_A \cdot g | Add step 5 and step 6 equations. |

8 | a = \frac{(m_B – m_A) \cdot g}{m_A + m_B} | Solve for acceleration $a$. |

Use the given number from the problem.

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

9 | a = \frac{(2.4 – 3.2) \cdot 9.8}{3.2 + 2.4} | Plug in known values. |

10 | a = \frac{-0.8 \cdot 9.8}{5.6} | Simplify the numerator and denominator. |

11 | a = -1.4 , \text{m/s}^2 | Calculate the acceleration. |

\boxed{a = -1.4 , \text{m/s}^2} Negative sign indicates downward direction.

**Part b: Calculate the tension in the string**

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

1 | T = m_A \cdot (g + a) | Tension equals mass A times (gravity plus acceleration). |

Using the previously calculated acceleration:

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

2 | T = 3.2 \cdot (9.8 – 1.4) | Plug in known values for mass and acceleration |

3 | T = 3.2 \cdot 8.4 | Simplify the terms in the parentheses. |

4 | T = 26.88 , \text{N} | Calculate the tension. |

**Part c: Calculate the final speed of mass A before it hits the ground**

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

1 | v = \sqrt{2 \cdot} | a |

2 | v = \sqrt{2 \cdot 1.4 \cdot 0.5} | Plug in the magnitude of $a$ and $=0.5m$. |

3 | v = \sqrt{1.4} | Calculate the expression under the square root. |

4 | v = 1.18 , \text{m/s} | Find the square root to get the final velocity. |

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- Statistics

Advanced

Conceptual

MCQ

A karate master is about to split a piece of wood with her hand. Select all she must do in order to deliver the maximum force to split the wood.

- Linear Forces, Momentum

Advanced

Mathematical

GQ

Find the downwards acceleration of an elevator, given that the ratio of a person’s stationary weight to their weight in the elevator is 5:4.

- Elevators, Linear Forces

Advanced

Mathematical

GQ

A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?

- Linear Forces, Rotational Motion, Torque

Advanced

Mathematical

MCQ

A ring is pulled on by three forces. If the ring is not moving, how big is the force F?

- Linear Forces, Tension

Beginner

Mathematical

GQ

A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?

- 1D Kinematics, Linear Forces

- 1.4 m/s
- 26.88 N
- 1.18 m/s

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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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