Kinetic Friction
Laws of Kinetic Friction
Learning Objectives:

Explain the physical nature of Kinetic Friction.
Friction is a force that always opposes the motion of an object. There are two types of friction static and kinetic. Kinetic friction occurs when two objects are moving against one another with some part of their surfaces in contact. Kinetic friction opposes the motion of the object and is proportional to the normal force acting on an object. The proportionality constant is the coefficient of friction, \(\mu_k\). The equation for this relationship is: \[F_k = \mu_k F_N\]
Solved Example: 29101
An empty cart is being rolled across a warehouse floor. If the cart was filled, the coefficient of kinetic friction between the cart and the floor would:
A. Decreases
B. Increases
C. Remain the same
D. Can't say
Ideally, the coefficient of kinetic friction only depends on the nature of the surfaces. It does not depend on any other factors, including the relative speed of the surfaces and the surface area of contact. It is also independent of normal reaction of the surface.
Correct Answer: C
Solved Example: 29102
Sand is often placed on an icy road because the sand:
A. Decreases the coefficient of friction between the tires of a car and the road
B. Increases the coefficient of friction between the tires of a car and the road
C. Decrease the gravitational force on a car
D. Increases the normal force of a car on the road
Unless the angle of the road is changed, the normal force on the car will NOT be affected, hence, option D is INCORRECT.
Sand CANNOT change the gravitational force. Hence, option C is also INCORRECT.
The keyword here is icy, means the sand is placed to avoid slipping of vehicles, hence, the coefficient of friction has to be increased.
Correct Answer: B
Motion Along an Inclined Plane
Learning Objectives:

Apply the concept of kinetic friction for single or multibody system.
Solved Example: 29201
A 50 kg block kept on a 15$^\circ$ inclined plane is pushed down the plane with an initial velocity of 20 m/s. If $\mu_k$ = 0.4 the determine the distance traveled by the block as it comes to rest. (Mumbai University Dec 2015, Engg Mechanics)
A. 160 m
B. 200 m
C. 320 m
D. 340 m
Net force parallel to the plane \begin{align*} F_{net} &= Mg \sin\theta  F_{FR}\\ &=Mg \sin\theta  \mu_k(Mg\cos \theta)\\ &= 50 \times 9.81 \times \sin 15^\circ  0.4 \times 50 \times 9.81 \times \cos 15^\circ\\ &= 126.95  189.51\\ &= 62.56\ N \end{align*} \begin{align*} a &= \dfrac{F_{net}}{M}\\ &= \dfrac{62.56}{50}\\ &= (1.251) m/s^2\\ \end{align*} Using equation of motion, \begin{align*} v^2 &= u^2 + 2as\\ (0)^2 &= (20)^2 + 2 (1.251)s\\ s &= \dfrac{400}{2\times 1.251} \approx 160\ m \end{align*}Correct Answer: A
Solved Example: 29202
A body of mass 25 kg resting on a horizontal table is connected by string passing over a smooth pulley at the edge of the table to another body of mass 3.75kg and hanging vertically. Initially the the friction between the mass A and the table is just sufficient to prevent the motion. If an additional 1.25kg is added to the 3.75kg mass, find the acceleration, in $m/s^2$ of the masses. Assume coefficient of kinetic friction to be 80% of the coefficient of static friction.
A. 0.65
B. 2
C. 2.5
D. 3
Hints:
Consider two cases separately. Draw free body diagrams. Write motion equation for each body separately.
Case I: When the blocks are at rest:
Tension in the string = 3.75 $\times$ 9.81 N
This tension is equal to limiting value of frictional force = $\mu_s N$
where, N = Normal reaction \[N = 25 \times 9.81 = 245.25 N\]
Solving,
Coefficient of static friction = 0.15
Coefficient of dynamic friction = 0.8 $\times$ 0.15 = 0.12
Case II: When the blocks are moving:
For hanging block: \[T' = 5 (ga)\]
For the block on the table: \[T'  \mu_k N = m_1 a\]
\[a = 0.654\ m/s^2\]
Correct Answer: A
Solved Example: 29203
A 10 kg mass block is initially pushed at a velocity of 1 m/sec along a horizontal track. If the block travels 10 meters before it stops, what is the coefficient of kinetic friction ($\mu_k$) between the block and track?
A. 0.005
B. 0.250
C. 0.450
D. 0.500
The initial kinetic energy K.E. is,
$\dfrac{1}{2}mv^2 =\dfrac{1}{2} (10 kg)(1 m/s)^2 = 5 J$
Work = force $\times$ distance, and work = energy.
The energy lost to friction = 5 J.
Force = energy/distance = (5 J)/(10 m) = 0.5 N.
Frictional force, \[F_f = \mu_k mg = \mu_k(10 kg)(9.81 m/s^2) = 98.1 \mu_k N \]Equating the forces, $0.5 N = 98.1 \mu_k N$,
$\mu_k = \dfrac{0.5}{98.1}= 0.005$
Correct Answer: A