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When you slide an object across a surface, such as a box across a floor, kinetic friction acts as a resistance between both surfaces. It works opposite to the direction of motion. The calculation for kinetic friction involves the coefficient of kinetic friction, denoted as \( \mu_k \), and the normal force \( N \). In this exercise, given \( \mu_k = 0.25 \), the formula is:

• \( f_k = \mu_k \times N \)

To find the normal force, we use the weight of the box, which is the gravitational force in a vertical direction. On horizontal surfaces, \( N = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity \( 9.8 \mathrm{m/s^2} \). Therefore,

• \( f_k = 0.25 \times 55 \mathrm{kg} \times 9.8 \mathrm{m/s^2} = 134.75 \mathrm{N} \)

Calculated kinetic friction then influences how much work is done by or against moving forces.

Normal force is the support force exerted by a surface perpendicular to the object resting on it. In our case, it's the force exerted upward by the floor on the box, balancing the downward gravitational force. It's crucial in calculating kinetic friction as it provides the \( N \) in \( f_k = \mu_k \times N \). For a box being pushed on a horizontal surface without vertical motion:

• Normal force \( N = mg \)
• \( m = 55 \mathrm{kg} \)
• \( g = 9.8 \mathrm{m/s^2} \)

This simplifies to \( N = 55 \mathrm{kg} \times 9.8 \mathrm{m/s^2} = 539 \mathrm{N} \). Since there is no vertical movement, normal and gravitational forces do no work because they are perpendicular to displacement (\( \theta = 90^\circ \)). Thus, work done by these forces is zero.

An applied force is what you exert to move the object across the surface. Here, it's the force \( \overrightarrow{\mathbf{P}} \) pushing the box forward, parallel to its direction of displacement. To compute the work done by this force, we use the formula:

• \( W = F \cdot d \cdot \cos \theta \)

Since \( \theta = 0^\circ \), the force follows the path of displacement directly, reducing \( \cos 0 = 1 \). Given the applied force \( F = 160 \mathrm{N} \) and displacement \( d = 7.0 \mathrm{m} \):

• \( W = 160 \mathrm{N} \times 7.0 \mathrm{m} \times 1 = 1120 \mathrm{J} \)

All the energy input through this force contributes to moving the box forward against friction.

Gravitational force is the force of attraction between the box and the Earth, commonly referred to as the box's weight. It acts downwards pulling the box towards the ground, calculated by:

• \( W = mg \)
• \( W = 55 \mathrm{kg} \times 9.8 \mathrm{m/s^2} = 539 \mathrm{N} \)

This force is crucial in context with normal force and kinetic friction. Gravitational force contributes to the normal force, thus affecting the kinetic friction, but does no work here since it acts perpendicular to the horizontal motion of the box (\( \theta = 90^\circ \)). That explains why its work calculation results in zero. Understanding these factors consolidates the role of gravitational force in analyzing movements across surfaces.