Problem 11 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15 and the push imparts an initial speed of \(3.5 \mathrm{~m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

The box will slide approximately 4.17 meters.

Step 1: Identify the Forces

First, identify the forces acting on the box. The major forces are the kinetic friction and the initial push. The force of kinetic friction is what slows the box down as it slides.

Step 2: Use the Formula for Frictional Force

Calculate the force of kinetic friction using the formula: \[ f_k = \mu_k \, N \] where \( \mu_k = 0.15 \) is the coefficient of kinetic friction, and \( N = mg \) is the normal force, with \( m \) being the mass and \( g = 9.8 \, \text{m/s}^2 \) being the acceleration due to gravity.

Step 3: Set Up the Equation of Motion

Apply Newton's second law and consider only the horizontal component, since the box is sliding horizontally. Using \[ F_{net} = ma = -f_k \] yields the following equation: \[ ma = -\mu_k mg \].Cancel \( m \) from both sides of the equation to obtain the acceleration: \[ a = -\mu_k g \]

Step 4: Calculate the Deceleration

Plug the given values into the formula for acceleration obtained:\[ a = -0.15 \times 9.8 = -1.47 \, \text{m/s}^2 \].This is the deceleration due to kinetic friction.

Step 5: Use Kinematic Equation to Find Distance

Now use the kinematic equation: \[ v^2 = u^2 + 2as \] where \( v = 0 \) m/s (final velocity when the box stops),\( u = 3.5 \) m/s (initial speed), and \( a = -1.47 \) m/s². Solve for \( s \):\[ 0 = (3.5)^2 + 2(-1.47)s \]

Step 6: Solve for Distance

Rearrange and solve for \( s \):\[ s = \frac{(3.5)^2}{2 \times 1.47} \approx 4.166 \].Thus, the box slides approximately 4.17 meters before coming to rest.

Key Concepts

Kinematic Equation

The kinematic equation is an essential tool utilized in physics to describe the motion of objects. These equations relate various motion parameters, such as velocity, acceleration, time, and displacement, which can provide crucial insights into the dynamics of moving objects.
The equation relevant to the exercise is given by:

• \( v^2 = u^2 + 2as \)

Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance traveled. In the context of our exercise, the box starts with an initial speed of 3.5 m/s and eventually comes to a stop. Thus, \( v = 0 \) m/s.

By substituting the known values into this kinematic equation, we can solve for \( s \), which represents the stopping distance of the box. This method is common in physics problems where objects are subjected to constant acceleration or deceleration, as it allows us to understand how far an object will travel before stopping.

Newton's Second Law

Newton's Second Law of Motion provides a fundamental principle for understanding how forces affect the motion of objects. It states that the net force acting on an object is equal to the product of its mass and acceleration:

• \( F_{net} = ma \)

This law helps us comprehend the box's motion in the problem, where the key force to consider is the kinetic friction that opposes the initial push.
The equation for the net force here includes the frictional force, thus leading to:

• \( ma = -f_k \)

Substituting \( f_k = \mu_k mg \) into the equation gives us:

• \( ma = -\mu_k mg \)

Canceling the mass \( m \) from both sides simplifies this to determine the acceleration due to friction, \( a = -\mu_k g \). This simplification is crucial as it shows that the deceleration depends only on the coefficient of friction and gravity, independent of mass.

Coefficient of Friction

The coefficient of friction is a dimensionless scalar that quantifies the frictional force between two surfaces in contact. It is critical for determining the deceleration caused by friction as an object moves across a surface.
Kinetic friction, a type of friction that acts on moving objects, is defined by:

• \( f_k = \mu_k N \)

where \( \mu_k \) represents the coefficient of kinetic friction, and \( N \) is the normal force. For horizontal surfaces, the normal force \( N \) equals \( mg \), the weight of the object.

In our exercise, \( \mu_k = 0.15 \), which tells us how much frictional resistance the floor offers against the sliding box. Understanding this value is crucial, as it directly affects the box's deceleration and therefore how far it will slide. A higher coefficient would lead to greater friction, resulting in a shorter sliding distance.