AP®PhysicsCollege BoardPhysics 1: Algebra-BasedRevision NotesUnit 2: Force & Translational DynamicsKinetic & Static FrictionKinetic Friction Force Equation

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

Written by: Ann Howell

Reviewed by: Caroline Carroll

Updated on 19 October 2024

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:

$| {\vec{F}}_{f, k} | = | μ_{k} {\vec{F}}_{n} |$

Where:
- $| {\vec{F}}_{f, k} | =$ magnitude of kinetic friction, measured in $N$
- $μ_{k} =$ coefficient of kinetic friction
- ${\vec{F}}_{n} =$ normal reaction force between the two contact surfaces, measured in $N$

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

Which of the following is the acceleration of the box?

A $1.33 m / s^{2}$

B $5.4 m / s^{2}$

C $16 m / s^{2}$

D $54 m / s^{2}$

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, $m = 12 kg$
Forward force, $F = 70 N$
Coefficient of kinetic friction, $μ_{k} = 0.450$

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

$| {\vec{F}}_{f, k} | = | μ_{k} {\vec{F}}_{n} |$

$| {\vec{F}}_{f, k} | = | μ_{k} {\vec{F}}_{g} | = | μ_{k} m g |$

$| {\vec{F}}_{f, k} | = 0.450 \cdot 12 \cdot 10$

$| {\vec{F}}_{f, k} | = 54 N$

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

${\vec{a}}_{s y s} = \frac{{\vec{F}}_{n e t}}{m_{s y s}}$

${\vec{a}}_{s y s} = \frac{70 - 54}{12}$

${\vec{a}}_{s y s} = 1.33 m / s^{2}$

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