Question

The coefficient of kinetic friction between a large box and the floor is \(0.21\). A person pushes horizontally on the box with a force of \(160 \mathrm{~N}\) for a distance of \(2.3 \mathrm{~m}\). If the mass of the box is \(72 \mathrm{~kg}\), what is the total work done on the box?

Short answer

The total work done on the box is approximately 26.84 Joules.

Step-by-step solution

Calculate the Normal Force

The first step is to calculate the normal force acting on the box. Since the box is on a flat surface and there are no vertical forces besides gravity and the normal force, the normal force can be calculated from the weight of the box. The weight is given by the force due to gravity, which is calculated as follows:\[ F_g = m \cdot g = 72 \text{ kg} \times 9.81 \text{ m/s}^2 = 706.32 \text{ N} \]Thus, the normal force \( F_N \) is equal to the gravitational force \( F_g \), so \( F_N = 706.32 \text{ N} \).

Calculate the Force of Friction

Now let's calculate the frictional force using the coefficient of kinetic friction. The force of friction \( F_f \) is calculated by multiplying the coefficient of kinetic friction \( \mu_k \) by the normal force \( F_N \):\[ F_f = \mu_k \cdot F_N = 0.21 \times 706.32 \text{ N} = 148.33 \text{ N} \]

Calculate the Net Force

The net force \( F_{net} \) acting on the box can be found by subtracting the frictional force from the applied force of 160 N:\[ F_{net} = 160 \text{ N} - 148.33 \text{ N} = 11.67 \text{ N} \]

Calculate the Work Done by the Net Force

The work done on an object is calculated using the formula:\[ W = F_{net} \times d \]where \( d \) is the distance over which the force is applied (2.3 m). Substituting the known values gives:\[ W = 11.67 \text{ N} \times 2.3 \text{ m} = 26.841 \text{ J} \]The total work done on the box, considering no vertical movement and friction, is approximately 26.84 Joules.

Key concepts

Kinetic Friction

Kinetic friction is a force that acts opposite to the direction of motion when two surfaces are sliding past one another.
It occurs whenever objects are in motion across a surface. In the case of our box moving across the floor, kinetic friction is a key factor affecting the work done on the box.
This force depends on two things:

• The nature of the surfaces in contact, which is represented by the coefficient of kinetic friction, \( \mu_k \).
• The normal force, which is the force perpendicular to the surfaces in contact.

This frictional force can be calculated with the equation:\[ F_f = \mu_k \times F_N \]In our example, where \( \mu_k = 0.21 \) and the normal force \( F_N = 706.32 \text{ N} \), the frictional force is \(148.33 \text{ N} \). This means that as the box slides, the kinetic friction works against the force trying to push the box forward.

Normal Force

The normal force is a crucial concept when it comes to understanding motion along surfaces. It is the force exerted by a surface to support the weight of an object resting on it.
In most scenarios like ours, this force acts perpendicular to the surface of contact.
For our box on a flat floor, the normal force is simply equal to the gravitational force acting on the box:\[ F_N = m \times g \]Where:

• \( m \) is the mass of the box, in our exercise, \(72 \text{ kg} \).
• \( g \) is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2 \).

Thus, we calculate the normal force as \(706.32 \text{ N} \).
This value becomes a part of determining other forces, such as the frictional force, enabling us to see how the box interacts with the surface beneath it.

Net Force

Net force is the sum of all the forces acting on an object. It determines the overall motion and direction the object will move.
In our specific exercise, two main forces are at play: the push force from the person and the kinetic friction pushing back.
To find the net force, we subtract the frictional force from the applied pushing force:\[ F_{net} = F_{applied} - F_f \]With:

• \( F_{applied} = 160 \text{ N} \), the force exerted by the person.
• \( F_f = 148.33 \text{ N} \), the opposing frictional force.

The net force computed is \( 11.67 \text{ N} \).
This force tells us that although some forces oppose the motion, there is still enough force pushing the box forward, resulting in its movement.

Work-Energy Principle

The work-energy principle is foundational in physics, linking the work done on an object to changes in its kinetic energy.
When a net force moves an object, work is done.
This principle helps us understand how energy is transferred from one form to another through work.For our scenario, work is calculated using:\[ W = F_{net} \times d \]Where:

• \( F_{net} \) is the net force acting on the box, \(11.67 \text{ N} \).
• \( d \) is the distance over which the force is applied, \(2.3 \text{ m} \).

Thus, the work done on the box is \( 26.841 \text{ J} \).
This work implies that energy equal to this value has been used to move the box over the given distance, showing how energy affects motion.
It's a crucial link between the concept of force and motion in physics.