Study GuidesExam QuestionsPast Papers
AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 2: Force & Translational DynamicsKinetic & Static FrictionKinetic Friction Force Equation

Kinetic Friction Force Equation (College Board AP® Physics 1: Algebra-Based)

Study Guide

Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Kinetic friction force equation

  • The magnitude of the kinetic friction force exerted on an object is the product of the normal force the surface exerts on the object and the coefficient of kinetic friction

  • The magnitude of kinetic friction is given by the equation:

open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals space open vertical bar mu subscript k F with rightwards arrow on top subscript n close vertical bar

  • Where:

    • open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals spacemagnitude of kinetic friction, measured in straight N

    • mu subscript k space equals spacecoefficient of kinetic friction

    • F with rightwards arrow on top subscript n space equals spacenormal reaction force between the two contact surfaces, measured in straight N

Kinetic friction equation

Diagram showing forces on an object: applied force, kinetic friction, gravitational force, and normal force. Directions and balance of forces indicated.
Kinetic friction is equal to the coefficient of friction multiplied by the normal reaction force

Coefficient of kinetic friction

  • The coefficient of friction does not have any units because it is a ratio of two forces between the two contact surfaces

    • the frictional force

    • the normal reaction force

  • The coefficient of kinetic friction depends on the material properties of the surfaces that are in contact

    • Rough surfaces like brick or concrete have a higher coefficient of kinetic friction compared to smooth surfaces like ice or plastic

Worked Example

A 12 space kg box is being pulled along a horizontal table by a parallel force of 70 space straight N. The coefficient of kinetic friction is 0.450.

Which of the following is the acceleration of the box?

A      1.33 space straight m divided by straight s squared

B      5.4 space straight m divided by straight s squared

C      16 space straight m divided by straight s squared

D      54 space straight m divided by straight s squared

The correct answer is A

Answer:

Step 1: Draw a free body diagram of the scenario

Diagram of a box on a surface with forces: 70N right, friction left, normal force up, gravitational force down. Coefficient of friction M=0.450.

Step 2: List the known quantities

  • Mass of box, m space equals space 12 space kg

  • Forward force, F space equals space 70 space straight N

  • Coefficient of kinetic friction, mu subscript k space equals space 0.450

Step 3: Substitute the known quantities into the equation for kinetic friction

open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals space open vertical bar mu subscript k F with rightwards arrow on top subscript n close vertical bar

open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals space open vertical bar mu subscript k F with rightwards arrow on top subscript g close vertical bar space equals space open vertical bar mu subscript k m g close vertical bar

open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals space 0.450 space times space 12 space times space 10

open vertical bar F with rightwards arrow on top subscript f comma k end subscript close vertical bar space equals 54 space straight N

Step 4: Substitute the known quantities into the equation for Newton's second law

a with rightwards arrow on top subscript s y s end subscript space equals space space F with rightwards arrow on top subscript n e t end subscript over m subscript s y s end subscript

a with rightwards arrow on top subscript s y s end subscript space equals space space fraction numerator 70 space minus space 54 over denominator 12 end fraction

a with rightwards arrow on top subscript s y s end subscript space equals space 1.33 space straight m divided by straight s squared

Last updated:

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Kinetic FrictionNext:Static Friction
    Author
    Reviewer
    Ann Howell

    Author: Ann Howell

    Expertise: Physics Content Creator

    Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.