A55-kg box is being pushed a distance of 7.0 m across the floor by a force 𝐏 whose magnitude is

step 1 This problem we have given a 55 kg box. Suppose this is box of mass 55 KG. It is pushed a distan

A55-kg box is being pushed a distance of 7.0 m across the...

A55-kg box is being pushed a distance of $7.0 \mathrm{m}$ across the floor by a force $\overrightarrow{\mathbf{P}}$ whose magnitude is $160 \mathrm{N}$. The force $\overrightarrow{\mathbf{P}}$ is parallel to the displacement of the box. The coefficient of kinetic friction is $0.25 .$ Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

Key Concepts

Work Done by a Force

Work is defined as the dot product of a force and displacement, expressed mathematically as W = F · d · cos(?), where ? is the angle between the force and displacement vectors. This concept is fundamental in analyzing how forces transfer energy to or from objects, depending on the alignment of the force with the displacement.

Kinetic Friction

Kinetic friction is the resistive force that acts opposite to the direction of an object’s motion when it slides over a surface. It is quantified by the product of the coefficient of kinetic friction and the normal force. This force always does negative work since it opposes the displacement, thereby reducing the kinetic energy of the object.

Force Directions and Components

Understanding the directional components of forces is crucial when calculating work. Forces that are parallel or anti-parallel to the displacement do work (positive if in the same direction, negative if opposite), whereas forces perpendicular to the displacement do no work, since the cosine of 90° is zero.

Normal and Gravitational Forces (Weight)

The normal force, exerted perpendicular to the contact surface, and the gravitational force (weight) are two key forces acting on an object in contact with a horizontal surface. In scenarios where displacement is horizontal, these forces do no work because their directions are perpendicular to the displacement.

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