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}}

Phy can also check your working. Just snap a picture!

- Statistics

Intermediate

Conceptual

MCQ

A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.

- Linear Forces, Tension

Beginner

Mathematical

MCQ

The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?

- Linear Forces

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

Beginner

Mathematical

GQ

What is weight of a person who has a mass of 75 kg?

- Linear Forces

Beginner

Mathematical

MCQ

The coefficient of kinetic friction is

- Friction, Linear Forces

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

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

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!

The most advanced version of Phy. Currently 50% off, for early supporters.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages
- Unlimited Image Uploads
- Unlimited Smart Actions
- Unlimited UBQ Credits
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- 200% Memory Boost
- 150% Better than GPT
- 75% More Accurate, 50% Faster
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

UBQ credits are specifically used to grade your FRQs and GQs.

You can still view questions and see answers without credits.

Submitting an answer counts as 1 attempt.

Seeing answer or explanation counts as a failed attempt.

Lastly, check your average score, across every attempt, in the top left.

MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.

Phy can give partial credit for GQs & FRQs.

Phy sees everything.

It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!

Understand you mistakes quicker.

For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.

Aim to increase your understadning and average score with every attempt!

10 Free Credits To Get You Started

*Phy Pro members get unlimited credits