Question

A force \(\mathrm{F}=\mathrm{kx}\) (where \(\mathrm{k}\) is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) From } \mathrm{x}=-4 \text { to } \mathrm{x}=-2 & \text { (P) Positive } \\ \hline \text { (b) From } \mathrm{x}=-2 \text { to } \mathrm{x}=-4 & \text { (Q) zero } \\ \hline \text { (c) From } \mathrm{x}=-2 \text { to } \mathrm{x}=+2 & \text { (R) negative } \\ \hline \end{array} $$ (A) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{Q}\) (B) \(a-P, b-Q, c-R\) (C) \(a-R, b-Q, c-P\) (D) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{R}\)

Short answer

The correct answer is (A): \(a-R, b-P, c-Q\)

Step-by-step solution

Analyze Case (a)

In this case, the particle is moved from \(x=-4\) to \(x=-2\). As the force and displacement are in opposite directions (the force is positive when \(x<0\) according to \(F=kx\), but the particle moves in the negative direction), the work done should be negative.

Analyze Case (b)

The particle moves from \(x=-2\) to \(x=-4\). As the force is positive and the displacement is also in the negative direction, the work done will be positive.

Analyze Case (c)

In this situation, the force and the displacement both change their signs as their starting point \(x=-2\) crosses the x=0 point to \(x=2\). This means that for half of the journey, the displacement and the force are in opposite directions, resulting in negative work, and for the other half of the journey, they are in the same direction, resulting in positive work. It is essential to understand that these two effects cancel each other out, resulting in zero work being done overall.

Match the Columns

According to the analysis done above, we can now match the columns: (a) from \(x=-4\) to \(x=-2\), the work done is negative, so it matches with R. (b) from \(x=-2\) to \(x=-4\), the work done is positive, so it matches with P. (c) from \(x=-2\) to \(x=2\), the work done is zero, so it matches with Q. So, the correct answer is (A): \(a-R, b-P, c-Q\)

Key concepts

Work done by a variable force

When dealing with a variable force, such as the force described by \( F = kx \), the work done is not simply the product of force and displacement. Instead, it requires calculus to determine the total work done as the force actually changes as the object moves through space.

The variable nature of the force means that it is dependent on the position \( x \). Calculating work done in such cases thus involves integrating the force over the path taken by the particle:

\[ W = \int F(x) \, dx = \int kx \, dx \]

This integration takes into account how the force changes at each point along the path, providing a complete picture of the work done over a specified interval. Understanding this concept allows us to accurately analyze situations where the force is not constant, making it key in physics problems involving springs, forces due to electric fields, or anything else that varies with position.

Direction of force and displacement

The direction of force and displacement plays a crucial role in determining the work done on an object. Work is calculated using the dot product of force and displacement vectors, mathematically expressed as:

\[ W = \vec{F} \cdot \vec{d} = |F||d| \cos \theta \]

where \( \theta \) is the angle between the force and displacement directions.

Here's what important to note:

• If the force and displacement are in the same direction (\( \theta = 0^\circ \)), \( \cos \theta = 1 \), and the work done is maximal and positive.
• If they are perpendicular (\( \theta = 90^\circ \)), \( \cos \theta = 0 \), no work is done.
• If the force and displacement are in opposite directions (\( \theta = 180^\circ \)), \( \cos \theta = -1 \), and the work done is negative.

The directionality between force and displacement images the physical concept of doing positive, negative, or no work, defining how energy is either transferred or remains unchanged.

Positive and negative work

The nature of positive and negative work is directly derived from the relative directions of force and displacement.

Positive work occurs when the force applied to an object by a force is in the same direction as the displacement of the object. This usually results in an increase in kinetic energy or movement of the object in line with the force applied.

In contrast, negative work happens when the force applied and the displacement are in opposite directions. Instead of adding energy to the object, negative work removes energy, often indicating a braking or resistant force at play.

These concepts are crucial for understanding real-world situations: positive work involves acceleration or extension, like when pressing the gas pedal in a car, while negative work might be applied when braking or encountering resistive forces like friction or air resistance.

Zero work

Zero work occurs in scenarios where the net work done by the force on an object is zero. This can happen under two primary conditions:

• The force and displacement are perpendicular. Imagine carrying a book while walking in a straight line; gravity acts downward, but the book moves horizontally, resulting in zero work with respect to gravity.
• The net work by various forces on the object cancels out. This is noticeable in case (c) of the exercise, where positive and negative work over two segments of the path nullify each other.

Zero work does not imply forces are absent, but simply that the energy transferred due to work is zero. It's essential in energy conservation assessments where forces counterbalance each other over time.

Force-displacement relationship

The force-displacement relationship demonstrates how the force changes relative to the position of an object.

In our scenario with \( F = kx \), the force is directly proportional to \( x \). This linear relationship means any displacement results in a directly proportional change in the force applied.

Key things to notice include:

• When \( x < 0 \), the force is negative, suggesting a restoring force direction. As \( x \) increases, crossing through zero, the force changes its direction as \( x \) becomes positive.
• On moving an object under such a force, the area under the force-displacement curve (a graphical representation) provides the work done. With changing force, understanding and integrating this relationship is vital for calculating work accurately.

This dynamic interplay between force and position underpins much of classical mechanics, dictating object motion in variable force fields.