Kinetic Friction - Definition, Types, Examples, FAQs

When we try to slide a body over a surface, the motion is opposed by a bonding between the body and the surface. This is caused by the electromagnetic forces between the charged particles at the surface and the body’s surface that is in contact. This resistance/opposition is represented by a force called friction. However, our topic of discussion is ‘Kinetic friction’ here. As we go further into the article, we will discuss what is kinetic friction and static friction, kinetic friction examples, kinetic friction formula, SI unit and dimensional formula of kinetic friction, coefficient of kinetic friction, work done by kinetic friction, differences between kinetic friction and static friction, and types of kinetic friction.

What is Kinetic Friction?

Kinetic friction definition: Friction is of two types- static and kinetic friction, also called dynamic friction. The type of friction that acts when a body slides over a surface is called kinetic friction denoted by $f_k$.

The two surfaces are moving relative to each other. The magnitude of kinetic friction force usually increases when the normal force increases. That is why it takes more force to slide a box full of books across the floor rather than to slide the box when it is empty.

This principle is also used in braking systems i.e. the harder the brake pads are squeezed against the rotating brake disks, the greater the braking effect produced.

We can also define kinetic friction as when two bodies in contact with each other move concerning each other, rubbing the surfaces in contact, the friction between them is kinetic. The direction of kinetic friction is the opposite of the velocity of the body moving on the surface of another body.

SI Unit and Dimensional Formula of Kinetic Friction

The SI unit of kinetic friction is Newton (N) or $\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2$

The dimensional formula of kinetic friction is,

$\left[f_k\right]=\left[{\mu }_k\right][N]=1\times \mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}$

$\left[f_k\right]=\mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}$

Coefficient of Kinetic Friction Formula

The magnitude of kinetic friction depends on the normal force or the normal reaction acting between the two bodies. So we can write-

$$f_k\propto N$$

$$f_k={\mu }_{k}N$$

where,

• $f_k$ is the kinetic frictional force
• ${\mu }_k$ is the coefficient of kinetic friction
• $N$ is the normal force

Where N is the normal force/normal reaction and ${\mu }_k$ is the proportionality constant used, which is called the coefficient of kinetic friction. The value of this coefficient of kinetic friction depends on the nature of the two surfaces in contact. The more slippery the surface is, the smaller the coefficient of kinetic friction i.e. if the surfaces are smooth, the coefficient is small and if the surfaces are rough, the value of the coefficient is large. Since this coefficient of kinetic friction is a quotient of two forces, it is a pure number and hence has no units.

Properties of Coefficient Kinetic Friction

• The coefficient of kinetic friction does not depend on the speed of the sliding bodies. However, this is true only for speeds that are not too large and are generally less than 10m/s.
• The coefficient of kinetic friction does not depend on the area of the surface in contact as long as the normal force N is the same.
• It is unitless and dimensionless.
• It depends on the nature and material of the two bodies in contact.
• It doesn't depend on the speed of the sliding bodies.

Kinetic friction examples include the friction that comes into play when anybody moves along a surface. We will discuss examples of kinetic friction alongside its different types.

Work Done by Kinetic Friction

As friction is force acting opposite to external force acting on the surface, this work done by a frictional force is nothing but a dot product of force with the displacement the body goes through.

$\mathrm{W}=-\mathrm{F}.\mathrm{ds}$

In terms of kinetic friction,

$W=f_k\cdot d\cdot \cos \theta$

Substitute $f_k={\mu }_{k}N$ in the work formula

$W={\mu }_{k}N\cdot d\cdot \cos \theta$

The angle $\theta$ between $f_k$ and $\overrightarrow{d}$ is ${180}^{\circ }$

So, $\cos \theta =\cos {180}^{\circ }=-1$

Substitute the value of cos, we get

$W={\mu }_{k}N\cdot d\cdot (-1)$

$W={\mu }_{k}N\cdot d\cdot (-1)$

Simplifying

$$W=-{\mu }_{k}Nd$$

What is Static Friction?

Imagine we are trying to slide a heavy almirah on a horizontal concrete surface by applying a force F. This concrete surface is not an ideal frictionless surface. So, if this force F is small, the almirah remains stationary and the force that resists or counteracts F, not allowing it to move, is called static friction denoted by $f_s$. If we increase the applied force, the almirah eventually starts moving. When the almirah is on the verge of slipping, the static force of friction is the maximum (limiting friction). At this point, F becomes greater than fs which is the maximum static force the surface can exert as a result the almirah breaks loose from the bonds between molecules and the surface of the floor and it begins to slide.

See figure-3. (a), (b), (c) shows that when there is no relative motion, the magnitude of static friction fs is less than or equal to ${\mu }_{s}N$, where s is the coefficient of static friction and N is the normal force. (d) shows that there is a relative motion when the magnitude of kinetic friction $f_k$ is equal to kN.

For a pair of given surfaces, the maximum value of $f_k$ depends on normal force and experiments show that this maximum value $f_k^{max}$ is proportional to N and the proportionality constant is ${\mu }_s$. In a given situation, the actual static friction can have any magnitude between zero and maximum given by ${\mu }_{s}N$.

Thus, ${\left(f_s\right)}_{max}={\mu }_{s}N$

Thus , $f_s\le {\mu }_{s}N$

This maximum value of static friction, $f_s^{max}$ is called limiting friction. Thus static friction can have any value between zero and its maximum or limiting value.

Differences Between Static and Kinetic/Dynamic Friction

Kinetic friction acts when there is a relative movement between the surfaces of 2 bodies or a body and floor. However, static friction acts on a body when it is at rest.

2. Kinetic friction is always comparatively less than limiting friction, i.e. $f_s>f_k$ hence, ${\mu }_s>{\mu }_k$

3. The kinetic friction does not depend on the magnitude of applied force. However, static friction depends on the magnitude of force applied.

4. The value of static friction can be anything from 0 to limiting friction and thus it can be zero but kinetic friction cannot have a zero value.

5. Static friction opposes motion n of a body when it’s at rest and then gets max when it starts to slide but kinetic friction is the opposition provided to an already moving body.

Derivation of Kinetic Energy Formula From Work-Energy Theorem

Work done is given as, $W=F\cdot d$

According to work energy theorem, $W_{\text{net }}=\mathrm{\Delta }KE$

$W_{\text{net }}=K{E}_{\text{final }}-K{E}_{\text{initial }}$

From Newton's second law we get, $F=m\frac{dv}{dt}$

Work done over a small displacement $dx$

$dW=Fdx$

Since $v=\frac{dx}{dt}$, we get $dx=vdt$

Thus, $dW=F\cdot vdt$

$dW=m\frac{dv}{dt}\cdot vdt$

$dW=mvdv$

Integrating total work done, $W={\int }_{v_0}^{v}mvdv$

$\begin{array}{rl}W& =m{\left[\frac{v^2}{2}\right]}_{v_0}^v\\ W& =\frac{m}{2}(v^2-v_0^2)\end{array}$

By the Work-Energy Theorem, the work done is equal to the change in kinetic energy

$$\mathrm{\Delta }KE=K{E}_{\text{final }}-K{E}_{\text{initial }}$$
Thus, the kinetic energy of the object at velocity $v$ is:

$$KE=\frac{1}{2}m{v}^2$$

Types of Kinetic Friction

There are three types of kinetic friction generally- i) sliding friction ii) rolling friction and iii) fluid friction. Let’s discuss them one by one-

1. Sliding friction- This is the type of kinetic friction that comes into play when an object slides over a surface. It is of course weaker than the static friction. The formula for sliding friction is the same as the kinetic friction where F is the force of sliding kinetic friction and the constant of proportionality is the coefficient of sliding kinetic friction. Examples are- when a block slides up or down an inclined plane, rubbing of hands together, rubbing of a matchstick, etc.

Rolling friction- A spherical body or a circular ring rolls over a horizontal surface without slipping and at every instant, the body touches the surface at just one point and this point does not move relative to the surface. The friction experienced by the rolling body is called rolling friction and it is much smaller than sliding friction hence very little force is required to keep rolling a body even if it is heavy, as compared to sliding it through the floor. That’s why the invention of wheels is a major milestone in the history of mankind.

To further keep things smoother, lubricants are used in ball the bearings of many machines which reduces kinetic friction to a great extent. Examples – a suitcase on wheels, cartwheels moving on the floor, a bicycle or any automobile with wheels, etc.

Fluid friction- Both the sliding and rolling friction come into play in the case of solids but in the case of liquids and gases, fluid friction comes into play. At least one of the surfaces is that of a fluid. It is the friction between the layers of a fluid that moves on top of each other or relative to each other. This internal opposition is also called viscosity. However if a body flows on top of a liquid or inside a fluid (which can be gas or a liquid), the body also experiences fluid friction.

This fluid friction experienced by the body can be minimized by giving a streamlined shape to the body. This is why airplanes, boats, and submarines are made streamlined. Nature has given birds and fishes streamlined bodies so that they can easily move through fluids. Some examples of fluid friction are-

Factors Affecting Kinetic Friction

1. The normal force
2. Coefficient of kinetic friction
3. Nature of the surface
4. Speed of motion
5. Temperature

Kinetic Energy Formula For Different Types of Motion

Recommended Topic Video

Solved Examples Based On Kinetic Friction

Question 1. A block rolled on a rough surface with a velocity of 8m/s comes to rest after traveling 4m. Compute the coefficient of friction ($\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$)

1) 0.8

2) 0.6

3) 0.4

4) 0.5

Solution

Given

The initial velocity of the block, u = 8 m/s,

Distance traveled before the block stops, s = 4 m

Let the coefficient of kinetic friction between the block and the surface be ${\mu }_k$.

F.B.D of the block-

From F.B.D

$$N=mg,f_k={\mu }_{k}N⟹f_k={\mu }_{k}mg$$
Assuming the acceleration of the block to be $a$,

$$f_k=ma⟹a=-{\mu }_{k}g$$
The block stops $(v=0)$ after traveling a displacement of 4 m.

Applying 3rd equation of motion-

$\begin{array}{c}{v}^2=u^2+2as\\ 0=u^2-2{\mu }_{k}gs⟹{\mu }_k=\frac{u^2}{2gs}\\ {\mu }_k=\frac{8\times 8}{2\times 10\times 4}=0.8\end{array}$

Question 2. The coefficient of static friction between two surfaces of the body depends

1) nature of surface

2) the shape of the surface in contact

3) area of contact

4) all of the above

Solution

Dependence of ($\mu$)

• Depends on material
• Nature of surfaces in contact

wherein

• It does not depend upon the apparent area of contact.

Friction force depends on the material and the nature of the surface in contact.

Hence, the answer is the Option (4).

Question 3. A rocket is fired vertically from the earth with an acceleration of $2g$, where $g$ is the gravitational acceleration. On an inclined plane inside the rocket, making an angle $\theta$ with the horizontal, a point object of mass $m$ is kept. The minimum coefficient of friction ${\mu }_{\text{min }}$ between the mass and the inclined surface such that the mass does not move is:

1) $\tan \theta$
2) $2\tan \theta$
3) $3\tan \theta$
4) $\tan 2\theta$

Solution

Kinetic or Dynamic Friction

$$f_k\propto R,f_k={\mu }_{k}R$$
Where:
$f_k=$ kinetic friction
${\mu }_k=$ coefficient of kinetic friction
$R=$ normal reaction force

$$f_k<f_s⟹{\mu }_k<{\mu }_s$$

${\mu }_k$ depends on the nature of the surfaces in contact.

Since the rocket is moving vertically upwards with an acceleration of 2g, the apparent acceleration experienced by the point object is g + 2g = 3g

Vertically downwards

where, $\begin{array}{c}N=3mg\cos \theta ,3mg\sin \theta =\mu (3mg\cos \theta )\\ \mu =\tan \theta \end{array}$

Hence, the correct option is 1.

Question 4. The upper half of an inclined plane with inclination $\varphi$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom of the coefficient of friction for the lower half is given by:

1) $2\tan \varphi$
2) $\tan \varphi$
3) $2\sin \varphi$
4) $2\cos \varphi$

Solution

Kinetic or Dynamic Friction -

$$f_k\propto R,f_k={\mu }_{k}R$$
Where:
$f_k=$ kinetic friction
${\mu }_k=$ coefficient of kinetic friction
$R=$ normal reaction force
- wherein $f_k<f_s⟹{\mu }_k<{\mu }_s$
${\mu }_k$ depends on the nature of the surfaces in contact.

For the upper half smooth incline, a component of g down the incline =$g\sin \varphi$

$v^2=2(g\sin \varphi )l_2$

For the lower half rough incline, frictional retardation =$\mu \mathrm{kg}\cos \varphi$

∴ Resultant acceleration $=g\sin \varphi -{\mu }_{k}g\cos \varphi$

$$\therefore 0=v^2+2(g\sin \varphi -{\mu }_{k}g\cos \varphi )l_2$$

or

$$0=2(g\sin \varphi )l_2+2g(\sin \varphi -{\mu }_k\cos \varphi )l_2$$

or

$$0=\sin \varphi +\sin \varphi -{\mu }_k\cos \varphi$$

or

$${\mu }_k\cos \varphi =2\sin \varphi$$

or

$${\mu }_k=2\tan \varphi$$

Hence, the correct option is 1.

Question 5. A smooth block is released at rest on a ${45}^{\circ }$ incline and then slides a distance d. The time taken to slide is n time as much to slide on a rough incline than on a smooth incline. The coefficient of friction is:

1) $\mu \mathrm{s}=1-1\mathrm{n}2$
2) $\mu \mathrm{s}=1-1\mathrm{n}2$
3) $\mu \mathrm{k}=1-1\mathrm{n}2$
4) $\mu k=1-1n2$

Solution

Kinetic or Dynamic Friction

$$f_k\propto R,f_k={\mu }_{k}R$$
Where:
$f_k=$ kinetic friction
${\mu }_k=$ coefficient of kinetic friction
$R=$ normal reaction force
Wherein $f_k<f_s⟹{\mu }_k<{\mu }_s$.
${\mu }_k$ depends on the nature of the surfaces in contact.
Component of $g$ down the plane=$g\sin \theta$

∴ For a smooth plane:

$$d=\frac{1}{2}(g\sin \theta )t^2\dots (\mathrm{i})$$
For rough planes:
Frictional retardation up the plane $={\mu }_k(g\cos \theta )$

∴$\begin{array}{c}d=\frac{1}{2}(g\sin \theta -{\mu }_{k}g\cos \theta )(nt{)}^2\\ \therefore \frac{1}{2}(g\sin \theta )t^2=\frac{1}{2}(g\sin \theta -{\mu }_{k}g\cos \theta )n^2{t}^2\end{array}$

or

$\sin \theta =n^2(\sin \theta -{\mu }_k\cos \theta )$
Putting $\theta ={45}^{\circ }$ :

$\sin {45}^{\circ }=n^2(\sin {45}^{\circ }-{\mu }_k\cos {45}^{\circ })$

or

$\frac{1}{\sqrt{2}}=n^2(\frac{1}{\sqrt{2}}-{\mu }_k\frac{1}{\sqrt{2}})$

Hence, the correct option is 3.

Also read:

• NCERT Exemplar Solutions for All Subjects
• NCERT Notes For All Subjects
• NCERT Solutions for All Subjects