Objective: Calculate the acceleration of the two objects and the tension in the string.

**For Acceleration**

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

1 | \text{Force on heavier mass, } F_1 = m_1g | Weight of the heavier object (49.0 kg). |

2 | \text{Force on lighter mass, } F_2 = m_2g | Weight of the lighter object (24.0 kg). |

3 | \text{Net force, } F_{\text{net}} = F_1 – F_2 | The difference in weights provides the net force. |

4 | F_{\text{net}} = m_1g – m_2g | Substitute the values of F_1 and F_2. |

5 | F_{\text{net}} = (m_1 – m_2)g | Factor out g. |

6 | a = \frac{F_{\text{net}}}{m_1 + m_2} | Newton’s second law, acceleration equals net force divided by total mass. |

7 | a = \frac{(m_1 – m_2)g}{m_1 + m_2} | Combine steps 5 and 6. |

8 | a = \frac{(49.0\text{ kg} – 24.0\text{ kg})(9.8\text{ m/s}^2)}{49.0\text{ kg} + 24.0\text{ kg}} | Substitute the masses and gravitational acceleration. |

9 | a = \frac{25.0\text{ kg} \times 9.8\text{ m/s}^2}{73.0\text{ kg}} | Simplify the equation. |

10 | a = 3.356\text{ m/s}^2 | Calculate to find acceleration. |

Final answer for acceleration: \boxed{a = 3.356\text{ m/s}^2}

**For Tension in the String**

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

1 | T – m_2g = m_2a | Newton’s second law for the lighter object. |

2 | T = m_2a + m_2g | Rearrange to solve for tension, T. |

3 | T = m_2(a + g) | Factor out m_2. |

4 | T = 24.0\text{ kg}(3.356\text{ m/s}^2 + 9.8\text{ m/s}^2) | Substitute the mass of the lighter object and calculated acceleration. |

5 | T = 24.0\text{ kg} \times 13.156\text{ m/s}^2 | Add a and g. |

6 | T = 315.744\text{ N} | Calculate to find tension. |

Final answer for tension: \boxed{T = 315.744\text{ N}}

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- Statistics

Intermediate

Mathematical

FRQ

A 10kg box is pushed to the right by an unknown force at an angle of 25° below the horizontal while a friction force of 50 N acts on the box as well. The box accelerates from rest and travels a distance of 4 m where it is moving at 3 m/s. Solve the following without the use of energy.

- 1D Kinematics, Linear Forces

Advanced

Mathematical

GQ

A rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?

- Linear Forces, Projectiles

Beginner

Mathematical

GQ

Two students push a 1750 kg car with a force of 758 N along a perfectly level road at a constant velocity of 4.00 m/s. Find the force of friction.

- Linear Forces

Advanced

Mathematical

GQ

A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.

- Linear Forces

Beginner

Mathematical

GQ

What force would have to be applied to start a 12.3 kg wood block moving on a surface with a static coefficient of friction of 0.438?

- Linear Forces

Intermediate

Proportional Analysis

GQ

A child on Earth has a weight of 500N. Determine the weight of the child if the earth was to triple in both mass and radius (3M and 3r).

- Gravitation, Linear Forces

Advanced

Mathematical

FRQ

_{3} = 1.0, m_{2} = 2.0, and m_{1} = 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above.

- Atwood Machine, Linear Forces, Multi-Body Systems

Intermediate

Mathematical

GQ

- Inclines, Linear Forces

Intermediate

Mathematical

FRQ

A train consists of 50 cars, each of which has a mass of 6.1 x 10^{3} kg. The train has an acceleration of 8.0 × 10^{-2} m/s^{2}?. Ignore friction and determine the tension in the coupling at the following places:

- Linear Forces, Multi-Body Systems

Intermediate

Mathematical

GQ

A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is 10.0°. What is the coefficient of kinetic friction between the sled and the hill’s surface?

- Inclines, Linear Forces

3.36 m/s^{2}, T = 316 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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