91Ó°ÊÓ

In this exercise, the applied force is a driving factor that propels the box across the floor. An applied force is any force that is exerted on an object by a person or another object. Here, it's represented by the symbol \( \overrightarrow{\mathbf{P}} \), with a magnitude of \( 160 \, \mathrm{N} \). Applied force is considered positive as it acts in the direction of the displacement.

When calculating work done by an applied force, remember to ensure the force direction is parallel to the displacement. This allows us to use the formula:

• Work \( W_P = P \times d \)

In this example, the work is given by \( 160 \, \mathrm{N} \times 7.0 \, \mathrm{m} = 1120 \, \mathrm{J} \).

Since the applied force moves the box in the same direction as the displacement, the work is positive.

Kinetic friction comes into play whenever two objects are in contact and moving relative to one another. This force, denoted as \( \overrightarrow{\mathbf{f_k}} \), opposes the motion. It acts in the opposite direction of displacement, contributing negative work. The magnitude of kinetic friction can be determined using the formula:

• \( f_k = \mu_k \times N \)

Here, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force. In our problem:

• \( \mu_k = 0.25 \)
• \( N = 539 \, \mathrm{N} \)

This results in a kinetic friction force of \( 134.75 \, \mathrm{N} \). The work done by this frictional force is negative:

• \( W_{f_k} = -f_k \times d \)

Therefore, the work is \(-134.75 \, \mathrm{N} \times 7.0 \, \mathrm{m} = -943.25 \, \mathrm{J} \).

It's crucial to note that kinetic friction always opposes the movement, hence the negative sign.

Gravitational force is the force that the Earth exerts on any object, pulling it towards its center. It acts downwards and is a product of mass and gravitational acceleration \( g \), calculated as \( F_g = mg \). Here, \( m = 55 \, \mathrm{kg} \) and \( g = 9.8 \, \mathrm{m/s^2} \), resulting in a force of \( 539 \, \mathrm{N} \).

In the context of this problem, the gravitational force doesn't contribute to the box's work in moving horizontally. This is because it acts perpendicular to the direction of displacement.

As a result, the work done by gravitational force is:

• \( W_g = 0 \, \mathrm{J} \)

Since there is no movement in the vertical direction, gravitational force does no work.

The normal force, \( \overrightarrow{\mathbf{N}} \), is a support force exerted by a surface, opposite to gravitational force and perpendicular to the surface. It ensures objects remain on surfaces without falling through. The normal force in this exercise equates to the gravitational pull, \( 539 \, \mathrm{N} \), when on a horizontal plane with no additional vertical forces.

Normal force typically does not perform any work if there’s no vertical displacement, as evidenced here where:

• Work done, \( W_N = 0 \, \mathrm{J} \)

This is because the normal force is vertical, not horizontal. The lack of vertical movement results in zero work by the normal force on a horizontally moving object.