Understanding Work Done by a Force

In physics, work done by a constant force is defined as the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the force and the displacement vectors. Mathematically, this is represented as:

\(W = \vec{F} \cdot \vec{S} = FS \cos\theta\)

Where:

• \(W\) is the work done.
• \(F\) is the magnitude of the applied force.
• \(S\) is the magnitude of the corresponding displacement.
• \(\theta\) is the angle between the force vector (\(\vec{F}\)) and the displacement vector (\(\vec{S}\)).

The sign of the work done depends on the value of \(\cos\theta\), which in turn depends on the angle between the force and the displacement.

Analyzing Force and Displacement Relationships for Work

Let's consider the different possibilities for the angle \(\theta\) between the applied force \(F\) and the corresponding displacement \(S\), and how they affect the work done:

• Positive Work: Work done is positive when the force has a component in the direction of displacement. This happens when \(0^\circ \le \theta < 90^\circ\). In this range, \(\cos\theta\) is positive, so \(W = FS \cos\theta\) is positive.
• Zero Work: Work done is zero when the force is perpendicular to the displacement. This happens when \(\theta = 90^\circ\). In this case, \(\cos 90^\circ = 0\), so \(W = FS \times 0 = 0\).
• Negative Work: Work done is negative when the force has a component opposite to the direction of displacement. This happens when \(90^\circ < \theta \le 180^\circ\). In this range, \(\cos\theta\) is negative, so \(W = FS \cos\theta\) is negative.

Evaluating the Given Options for Negative Work

Let's examine each option provided in the question based on our understanding of work and the angle \(\theta\):

1. Perpendicular to each other: If the applied force \(F\) and the corresponding displacement \(S\) are perpendicular, the angle \(\theta\) between them is \(90^\circ\).

   \(W = FS \cos(90^\circ) = FS \times 0 = 0\)

   This results in zero work, not negative work.

2. Parallel to each other: If the applied force \(F\) and the corresponding displacement \(S\) are parallel, the angle \(\theta\) between them is \(0^\circ\).

   \(W = FS \cos(0^\circ) = FS \times 1 = FS\)

   Since \(F\) and \(S\) are magnitudes (always non-negative), this results in positive work (assuming \(F>0\) and \(S>0\)).

3. Anti-parallel to each other: If the applied force \(F\) and the corresponding displacement \(S\) are anti-parallel, they are in opposite directions. The angle \(\theta\) between them is \(180^\circ\).

   \(W = FS \cos(180^\circ) = FS \times (-1) = -FS\)

   Since \(F\) and \(S\) are magnitudes (always non-negative), this results in negative work (assuming \(F>0\) and \(S>0\)).

4. Equal in magnitude: The magnitudes of force and displacement being equal (\(F=S\)) tells us nothing about the direction between them. The angle \(\theta\) could be anything. For example, if \(F=S\) and they are parallel, work is \(F^2\) (positive). If \(F=S\) and they are anti-parallel, work is \(-F^2\) (negative). If \(F=S\) and they are perpendicular, work is \(0\). This condition alone does not determine the sign of the work done.

Conclusion on Negative Work

Based on the analysis of the options, a negative work is done when the applied force \(F\) and the corresponding displacement \(S\) are anti-parallel to each other. This means the force acts in the direction opposite to the motion.

Examples of negative work include the work done by friction on a moving object, or the work done by gravity on an object being lifted upwards.

Revision Table: Work, Force, and Displacement

Additional Information: Work-Energy Theorem

The work-energy theorem is a fundamental concept related to work done by forces. It states that the net work done on an object by all forces is equal to the change in its kinetic energy (\(\Delta KE\)).

\(W_{net} = \Delta KE = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{i}^2\)

Where:

• \(W_{net}\) is the total work done by all forces.
• \(m\) is the mass of the object.
• \(v_f\) is the final speed.
• \(v_i\) is the initial speed.

When negative work is done on an object (e.g., by friction slowing it down), the net work is often negative, resulting in a decrease in kinetic energy and thus a decrease in speed. Conversely, positive work increases kinetic energy and speed.