Find Coefficient Of Friction Given Velocity And Distance: Detailed Analysis and Problems

To find the coefficient of friction, the normal reaction and the friction involved in the necessary quantities. But how to find coefficient friction given velocity and distance

The velocity and the distance contribute to the friction evolving between the surfaces. The impact of these two quantities can be resolved. By knowing the velocity and the distance moved by the object, the coefficient of friction can be calculated.

How to find coefficient of friction given velocity and distance moved by the object

To find the coefficient of friction given velocity and distance, let us consider an object of mass ‘m’ moving with a velocity ‘v’ at a distance ‘d’ from the initial position. The friction force Ff is retards the motion of the object in the direction opposite to the movement. The normal reaction Fn is acting perpendicular to the motion of the object. The motion of the object is influenced by the gravity ’g’, which results in a normal reaction.

Find Coefficient Of Friction Given Velocity And Distance
To find coefficient of friction given velocity and distance

We can find the coefficient friction given velocity and distance by two methods. Let us it discuss one by one.

Method 1: considering the work done by the friction

The work done by the friction on the object is given by

W = PE+ KE+ Eloss 

Since the object is moving, the stored potential energy is zero, and the energy loss is negligible during the process. So the work done can be rewritten as

CodeCogsEqn 24

Where m is the mass of the object and v is the velocity at which the object is moving.

This work done is equal to the friction force times the distance, so we can write the equation as

CodeCogsEqn 27
CodeCogsEqn 29

The friction force acting on the object is given by

Ff = µFn

Where µ is the coefficient of friction and Fn is the normal reaction.

The normal reaction is equal to the net weight of the object given by Fn=mg

So that the friction is given by the equation,

Ff = µmg …..(2)

Since the above two equations of friction are the same, we can equate them

CodeCogsEqn 30

Rearranging the equation we get,

CodeCogsEqn 31

Now consider that initially the object is moving with velocity v0, with the time its velocity changes and finally it is moving with velocity vf covering the distance d, then the coefficient of friction is given by

CodeCogsEqn 33

Method 2: By using the Kinematics

The kinematic equation of motion for given velocity and distance is,

vf2 = v02 + 2ad

Where, vf2 is the final velocity of the moving object.

v02 is the initial velocity of the moving object.

a is the acceleration, and d is the distance travelled by the object.

Here we consider the object is moving with a constant velocity so that the final velocity vf2 become zero. Hence we can modify the equation as

0 = v02 + 2ad

-v02 = 2ad

CodeCogsEqn 34

From Newton’s second law of motion, the relation between the acceleration and the force is given by,

F = m*a

Substituting in the above equation, we get

CodeCogsEqn 35

The force acting on the object cannot have a negative value. We can take the magnitude of the equation as,

CodeCogsEqn 36
CodeCogsEqn 37

Since in this case, we have considered the force acting on the object is friction force so that substituting the formula of the friction, we get the equation as

CodeCogsEqn 38

Generally, we can write the equation as

CodeCogsEqn 39

On rearranging the terms, we get the coefficient of friction as,

CodeCogsEqn 40

It is clear that the coefficient of friction arrived from both methods are the same. Using the above formula, we can find the coefficient of friction given velocity and distance.

Solved Problems On Coefficient Of Friction

An object of mass of 2kg is moving with a velocity of 12ms-2, the friction force acting on the body make the object stop at a distance of 22m. Find the coefficient of friction and hence calculate the friction force. The acceleration due to gravity g given as 10ms-2.

Solution:

               Given: Mass of the object m = 2kg

                            Velocity v = 13ms-2

                            Distance covered by the object d = 22m

                            Acceleration due to gravity g = 10ms-2

The formula to find coefficient of friction given velocity and distance is

CodeCogsEqn 31

Substituting the values of the given terms in the above equation

CodeCogsEqn 41
CodeCogsEqn 42

µ = 0.38

The formula to calculate friction is

Ff = µmg

Ff = 0.38×10×2

Ff = 7.68N.

Find the coefficient of friction given velocity and distance as 28ms-2 and 34m respectively and hence find the normal reaction and the friction force. (Given: Mass of the object is 4kg and Acceleration due to gravity is 10 ms-2).

Solution:

The velocity is 17ms-2

The distance travelled by the object is34m. 

The coefficient of friction for given velocity and distance is given by the formula

CodeCogsEqn 31

Substituting the values in the expression,

CodeCogsEqn 44
CodeCogsEqn 43

µ = 0.425

The normal reaction is given by FN = m*g

FN = 4 × 10

FN = 40 N.

The friction force acting on the object is Ff = µ FN

Ff = 0.425 × 40

Ff = 17 N.

A body of mass of 12kg is moving on a rough surface. It travelled a distance of 72m, and then its motion is hindered by the friction force of 45N. Calculate the coefficient of friction and hence find the velocity at which the body is moving.

Solution:

Mass of the body m = 12kg

The distance traveled by the body d = 72m

The friction force acting on the body Ff = 45N

Acceleration due to gravity g = 9.8 ms-2.

The coefficient of friction for given friction is given by the formula

CodeCogsEqn 45

Substituting values in the above equation

CodeCogsEqn 46
CodeCogsEqn 47

µ = 0.382

To find the velocity, let us consider the equation

CodeCogsEqn 31

On rearranging the terms, we get the equation for the velocity as,

v2 = 2µgd

Substituting the values

v2 = 2× 0.382× 9.8× 72

v2 = 539.07

Taking the square root, we get

CodeCogsEqn 49

The velocity at which the body is moving is v = 23.21 ms-2.

The coefficient of friction is 0.46, and the mass of the object is 7kg. The object is moving with a constant velocity of 46 ms-2. Calculate the distance travelled by the object after friction retards the object’s motion.

Solution:

The coefficient of friction µ = 0.46

Mass of the object m = 7kg

Velocity of the object v = 16 ms-2

Acceleration due to gravity g = 9.8 ms-2

CodeCogsEqn 31

Rearranging the expression

CodeCogsEqn 50
CodeCogsEqn 51
CodeCogsEqn 52

d = 28.39m.

The object covers a distance of 28.39m before it stops its motion.

A block of mass 5kg is moving with the initial velocity of 12ms-2. After a time t, its velocity is increased by 19ms-2 and covers a distance of 33m, then its motion is stopped by the friction. Find the coefficient of friction given velocity and distance and hence find the friction required to stop the motion of the object.

Solution:

The initial velocity of the object v0 = 12 ms-2

The final velocity of the object vf = 19ms-2

The distance covered by the object d = 33m

Mass of the object m = 5kg

The coefficient of friction for given initial and final velocity is given by the expression

CodeCogsEqn 33
CodeCogsEqn 53
CodeCogsEqn 55

µ = 0.33

The friction required to stop the motion of the object is

Ff = µmg

Ff = 0.33× 5× 9.8

Ff = 16.17 N.

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