Kinetic Friction: Videos & Practice Problems

Kinetic friction, denoted as f_k, is a resisting force that occurs when two rough surfaces slide against each other. It acts opposite to the direction of motion, with its magnitude calculated using the equation f_k = μ_k × ⁢N, where μ_k is the coefficient of kinetic friction and N is the normal force. The coefficient varies between 0 (smooth surfaces) and 1 (rough surfaces), influencing the frictional force experienced during motion.

Kinetic Friction Problems

Kinetic friction, denoted as \( f_k \), is a resisting force that occurs when two rough surfaces slide against each other. This force is crucial in understanding motion, as it opposes the direction of velocity. For instance, when a book slides across a table, kinetic friction acts in the opposite direction to the motion, ultimately bringing the book to a stop. The direction of kinetic friction is always opposite to the velocity vector, which is essential for analyzing motion in various scenarios, such as a book sliding down an incline where friction acts upward along the ramp.

The magnitude of kinetic friction can be calculated using the formula:

\( f_k = \mu_k \cdot N \)

In this equation, \( \mu_k \) represents the coefficient of kinetic friction, a unitless value that quantifies the roughness of the surfaces in contact. This coefficient ranges from 0 to 1, where 0 indicates perfectly smooth surfaces with no friction, and values closer to 1 indicate rougher surfaces with higher resistance. For example, ice has a low coefficient of kinetic friction, while materials like cinder blocks have a high coefficient.

To illustrate the application of kinetic friction, consider a 10-kilogram box sliding on a flat surface at a velocity of 2 meters per second, with a coefficient of kinetic friction of 0.4. To find the kinetic friction force, we first determine the normal force \( N \), which, for a flat surface, is equal to the weight of the box:

\( N = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \)

Substituting this into the kinetic friction formula gives:

\( f_k = 0.4 \cdot 98 \, \text{N} = 39.2 \, \text{N} \)

This friction force acts to the left, opposing the motion of the box.

Next, to find the acceleration of the box, we apply Newton's second law, \( F = ma \). Since the only horizontal force acting on the box is the kinetic friction, we set up the equation:

\( -f_k = ma_x \)

Substituting the known values, we have:

\( -39.2 \, \text{N} = 10 \, \text{kg} \cdot a_x \)

Solving for acceleration yields:

\( a_x = -3.92 \, \text{m/s}^2 \)

The negative sign indicates that the acceleration is in the opposite direction of the velocity, consistent with the effect of kinetic friction slowing the box down.

Pushing a 10-kg toolbox across the floor, you find that the box moves at a constant speed when you push horizontally with a force of 39 N. What is the coefficient of kinetic friction between the floor and the toolbox?

• A. 0.2
• B. 0.4
• C. 2.5

Answer: (B)

You push on a 3-kg box to give it an initial speed of 5 m/s across a floor. If μ_k = 0.3, how far does the box travel before coming to a stop?

• A. 8.6m
• B. 2.9m
• C. 7.7m
• D. 4.3m

Answer: (D)

Pushing Down on a Block with Friction

In this problem, we analyze a 20-kilogram box moving along the floor while experiencing a downward force of 30 newtons. The goal is to determine the horizontal force required to maintain a constant velocity of 2 meters per second. To solve this, we start by constructing a free body diagram that illustrates all the forces acting on the box.

The forces include the gravitational force acting downward, represented as \( F_g = mg \), where \( m \) is the mass (20 kg) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²). The downward force applied to the box is \( F_{\text{down}} = 30 \, \text{N} \). Additionally, there is a normal force \( N \) acting upward and a kinetic friction force \( F_k \) opposing the motion to the left.

Since the box is moving at a constant speed, the acceleration is zero, indicating that the net force in the horizontal direction must also be zero. This leads us to the equation:

\[ \sum F_x = F - F_k = 0 \]

From this, we can deduce that the applied force \( F \) must equal the kinetic friction force \( F_k \). The equation for kinetic friction is given by:

\[ F_k = \mu_k \cdot N \]

where \( \mu_k \) is the coefficient of kinetic friction, which is provided as 0.3. To find the normal force \( N \), we analyze the vertical forces:

\[ \sum F_y = N - mg - F_{\text{down}} = 0 \]

Rearranging this gives:

\[ N = mg + F_{\text{down}} \]

Substituting the values, we find:

\[ N = (20 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) + 30 \, \text{N} = 196 \, \text{N} + 30 \, \text{N} = 226 \, \text{N} \]

Now, substituting \( N \) back into the kinetic friction equation gives:

\[ F_k = 0.3 \cdot 226 \, \text{N} = 67.8 \, \text{N} \]

Thus, the horizontal force required to keep the box moving at a constant speed of 2 meters per second is 67.8 newtons. This analysis illustrates the balance of forces in a system at equilibrium, emphasizing the importance of understanding both horizontal and vertical forces in mechanics.

Here’s what students ask on this topic:

What is kinetic friction and how is it different from static friction?

Kinetic friction, denoted as f_k, is a resisting force that occurs when two rough surfaces slide against each other. It acts opposite to the direction of motion. The magnitude of kinetic friction is calculated using the equation $f$_k=μ_k×N, where $\mu$_k is the coefficient of kinetic friction and $N$ is the normal force. Static friction, on the other hand, is the force that prevents two surfaces from sliding past each other. It is generally higher than kinetic friction and must be overcome to initiate motion. Once motion starts, kinetic friction takes over.

How do you calculate the force of kinetic friction?

The force of kinetic friction, f_k, is calculated using the equation $f$_k=μ_k×N. Here, $\mu$_k is the coefficient of kinetic friction, which is a measure of how rough the two surfaces in contact are, and $N$ is the normal force, which is the perpendicular force exerted by a surface on an object in contact with it. The coefficient of kinetic friction is a unitless number between 0 and 1, with higher values indicating rougher surfaces.

What factors affect the coefficient of kinetic friction?

The coefficient of kinetic friction, $\mu$_k, is influenced by the nature of the surfaces in contact. Rougher surfaces have higher coefficients, while smoother surfaces have lower coefficients. For example, ice on ice has a low coefficient of kinetic friction, close to 0, while rubber on concrete has a higher coefficient, closer to 1. The coefficient is a property of the materials involved and does not depend on the surface area or the speed of sliding.

How does kinetic friction affect the motion of an object?

Kinetic friction acts opposite to the direction of motion, thereby resisting the movement of an object. It causes the object to slow down and eventually stop if no other forces are applied. The magnitude of kinetic friction is given by $f$_k=μ_k×N. This force is crucial in everyday activities, such as walking or driving, where it provides the necessary resistance to control motion. Without kinetic friction, objects would continue to move indefinitely once set in motion.

Can the coefficient of kinetic friction be greater than 1?

In most practical scenarios, the coefficient of kinetic friction, $\mu$_k, ranges between 0 and 1. However, in some cases involving very rough surfaces or special materials, it can exceed 1. This indicates extremely high resistance to sliding. For example, certain rubber materials on rough surfaces can have coefficients slightly greater than 1. Nonetheless, such instances are rare, and for most common materials, the coefficient remains within the 0 to 1 range.