Problem 31 - College Physics

Question

An \(85 \mathrm{~N}\) box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of \(0.90 \mathrm{~m} / \mathrm{s}\) each second. The push force has a horizontal component of \(20 \mathrm{~N}\) and a vertical component of \(25 \mathrm{~N}\) downward. Calculate the coefficient of kinetic friction between the box and floor.

Step-by-Step Solution

Answer

The coefficient of kinetic friction is approximately 0.111.

Step 1: Understand the Forces

The forces acting on the box include its weight (85 N acting downwards), the push force with a horizontal component of 20 N and a vertical component of 25 N downwards, and the kinetic friction force opposing the motion.

Step 2: Determine Normal Force

Calculate the normal force. Since two forces act downwards (weight and vertical push), the normal force (F_N) will counteract these. \[ F_N = ext{Weight} + ext{Vertical Component} = 85 \, \text{N} + 25 \, \text{N} = 110 \, \text{N} \]

Step 3: Apply Newton’s Second Law

Write Newton's second law for horizontal motion: the net force equals mass times acceleration.\[ F_{ ext{net}} = m imes a \] The net force can also be given by the difference between the horizontal push and kinetic friction. Let the mass (m) be \( F_g/g = 85/9.81 \approx 8.66 \, \text{kg} \). The box decelerates at 0.90 \, \text{m/s}^2.

Step 4: Calculate Net Force and Friction

Calculate the net force using \( F_{ ext{net}} = m \times a \). Substituting the values:\[ F_{ ext{net}} = 8.66 \, \text{kg} \times 0.90 \, \text{m/s}^2 = 7.794 \, \text{N} \]Since the net force is the difference between the horizontal component of the push and the friction force: \[ f_k = 20 \, \text{N} - 7.794 \, \text{N} = 12.206 \, \text{N} \]

Step 5: Calculate the Coefficient of Kinetic Friction

Use the definition of kinetic friction force, where \( f_k = \mu_k \times F_N \).Rearrange to solve for \( \mu_k \):\[ \mu_k = \frac{f_k}{F_N} = \frac{12.206 \, \text{N}}{110 \, \text{N}} \approx 0.111 \]

Key Concepts

Newton's Second Law

Newton's second law is the cornerstone for understanding how forces impact motion. It states that the net force acting on an object is equal to the product of its mass and acceleration: \( F_{\text{net}} = m \times a \). In our exercise, the box of oranges is experiencing a deceleration due to friction. To find out how the forces interact, we need to calculate the net force and set it equal to mass times acceleration. Here, the mass can be derived from the weight by dividing by the gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \). For the box, the mass is approximately 8.66 kg.

• Net force: Difference between the applied horizontal push and friction.
• Box's deceleration: \( 0.90 \, \text{m/s}^2 \).

By plugging these values into the equation, we gain a clearer perspective of the interplay of forces slowing the box down.

Normal Force Calculation

Normal force, an essential aspect in this problem, is the support force exerted by a surface perpendicular to an object resting on it. For horizontal surfaces, the normal force usually equals the object's weight. However, here, the push applies an additional vertical force on the box. The weight of the box is 85 N, and along with the vertical push component of 25 N downward, both influence the normal force. Calculating the normal force involves summing these vertical forces, which the surface must counterbalance:\[ F_N = 85 \, \text{N} + 25 \, \text{N} = 110 \, \text{N} \]

• Normal force: Helps determine the frictional force acting on the box.
• Increased downward force: Results in a higher normal force than the weight alone.

Understanding the normal force's role is crucial for integrating it with frictional calculations.

Deceleration Due to Friction

Deceleration due to friction is a common effect when an object slows down on a surface. The frictional force is a product of the coefficient of kinetic friction and the normal force: \( f_k = \mu_k \times F_N \). Friction opposes the motion, decelerating the box, and its magnitude is essential for calculating the coefficient of kinetic friction.From the given exercise:

• Horizontal push force: 20 N.
• Net force resulting from deceleration: 7.794 N.
• Kinetic friction \( f_k \): 12.206 N.

We determine \( \mu_k \) by dividing the frictional force by the normal force. Solving gives:\[ \mu_k = \frac{12.206 \, \text{N}}{110 \, \text{N}} \approx 0.111 \]

• Smaller \( \mu_k \): Indicates less resistance from the surface.
• Deceleration explained: Directly linked to the frictional force.

Deciphering friction's role allows us to understand the link between forces, motion, and surface interaction.