Understanding Work Done by Gravity

The question asks to identify the statement that is NOT correct concerning the work done by gravity on a body. To solve this, we need to recall the properties of gravitational force and how work is calculated in physics.

Analyzing Gravity as a Conservative Force

Gravity is classified as a conservative force. Conservative forces possess specific characteristics related to the work they perform:

• Path Independence: The work done by a conservative force moving an object between two points is solely determined by the object's initial and final positions. It does not depend on the trajectory or path taken between these points.
• Potential Energy Association: Conservative forces are associated with a potential energy function. The work done by the force is equal to the negative change in the potential energy ($W_{cons} = -\Delta PE$).
• Zero Work on Closed Paths: If an object moves along a path and returns to its starting point (a closed path), the net work done by a conservative force over that entire path is zero.

The general formula for work done is $W = \vec{F} \cdot \vec{d}$, where $\vec{F}$ is the force and $\vec{d}$ is the displacement.

Evaluating Each Statement about Work Done by Gravity

Let's analyze each option provided:

Statement 1: Work done is independent of the path followed by the body

This aligns perfectly with the definition of a conservative force. Since gravity is conservative, the work it does is indeed independent of the path. Thus, this statement is correct.

Statement 2: Work done depends only on the vertical distance separating the initial and the final position

The force of gravity acts vertically downwards. Let the gravitational force be $\vec{F}_g = -mg\hat{j}$ (where $m$ is mass, $g$ is acceleration due to gravity, and $\hat{j}$ is the unit vector in the vertical direction). The displacement can be represented as $\vec{d} = \Delta x \hat{i} + \Delta y \hat{j}$ (where $\Delta x$ is horizontal displacement and $\Delta y$ is vertical displacement). The work done by gravity ($W_g$) is:

$W_g = \vec{F}_g \cdot \vec{d} = (-mg\hat{j}) \cdot (\Delta x \hat{i} + \Delta y \hat{j})$

Using the properties of dot products ($\hat{j} \cdot \hat{i} = 0$ and $\hat{j} \cdot \hat{j} = 1$), we get:

$W_g = -mg\Delta y$

This formula clearly shows that the work done by gravity ($W_g$) depends only on the vertical displacement ($\Delta y$), which represents the vertical distance between the initial and final positions. Therefore, this statement is correct.

Statement 3: Work done depends on the path followed by the body

This statement is in direct contradiction to the properties of conservative forces. As we've established, gravity is conservative, and the work done by it is path independent. Therefore, this statement claiming dependence on the path is NOT correct.

Statement 4: Work done is zero if the body is displaced horizontally to the force of gravity

The force of gravity is vertical. If a body moves horizontally, its displacement vector is perpendicular to the force of gravity vector. The dot product of two perpendicular vectors is zero.

Mathematically, if $\vec{F}_g$ is vertical and the displacement $\vec{d}$ is horizontal, then $\vec{F}_g \perp \vec{d}$. The angle between them is $90^\circ$. Since $W_g = |\vec{F}_g| |\vec{d}| \cos(90^\circ)$ and $\cos(90^\circ) = 0$, the work done is zero.

Therefore, this statement is correct.

Conclusion on the Incorrect Statement

By analyzing the properties of gravity as a conservative force and applying the definition of work, we find that the statement "Work done depends on the path followed by the body" is the one that is NOT correct.