A force F = -k(y i^ + x j^) (where, k is a positive constant) acts on a particle moving in the x-y p

Q: Why is the work done on the first path exactly zero?

On the first path (along the x-axis), the y-coordinate is zero. If you plug $y=0$ into the force equation, the force becomes purely vertical ($-kx\hat{\mathbf{j}}$). However, the particle is moving horizontally. Since the force is perpendicular to the displacement, the dot product is zero, meaning no work is done.

Q: What does an 'exact differential' mean physically in this context?

When the work differential $dW$ can be written as the exact differential of some scalar function (like $d(xy)$), it means the force field is conservative. The work done depends only on the starting and ending coordinates, not on the specific path taken between them. The scalar function is closely related to the potential energy.

Q: Could we have taken a straight diagonal path from $(0,0)$ to $(a,a)$ to get the answer?

Yes, absolutely! Because the force is conservative (as proven by the exact differential), any path will give the same result. For a straight diagonal line, the equation is $y=x$. You could substitute $y=x$ and $dy=dx$ into the work integral, integrate from $x=0$ to $x=a$, and you would still get $-ka^2$.

Q: Why does the x-component of the force not do any work on the second path?

On the second path, the particle is moving straight up, parallel to the y-axis. Its displacement vector only has a $\hat{\mathbf{j}}$ component. The x-component of the force points horizontally ($\hat{\mathbf{i}}$). A horizontal force cannot do work on an object that is only moving vertically, because the angle between them is 90 degrees ($\cos 90^\circ = 0$).

Q: Is this force conservative? How can we tell?

Yes, it is conservative. There are two main ways to tell here. First, we showed that the work done $dW$ is an exact differential, which is a hallmark of conservative forces. Second, if you calculate the curl of the force vector ($\nabla \times \mathbf{F}$), you will find it equals zero, which is the mathematical test for a conservative force field.

Question

A force $\mathbf{F} = -k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}})$ (where, $k$ is a positive constant) acts on a particle moving in the $x$-$y$ plane. Starting from the origin, the particle is taken along the positive $X$-axis to the point $(a, 0)$ and then parallel to the $Y$-axis to the point $(a, a)$. The total work done by the force $\mathbf{F}$ on the particle is

Options:

(A) $-2ka^2$

(B) $2ka^2$

(C) $-ka^2$

(D) $ka^2$

Sigma Percentile · Accepted Answer

### The Physics of Variable Forces

When dealing with constant forces, calculating work is as simple as taking the dot product of the force vector and the displacement vector: $W = \mathbf{F} \cdot \mathbf{s}$. However, the universe is rarely that simple. In many physical scenarios, the force acting on an object changes as the object moves through space. This is known as a **variable force**.

To calculate the work done by a variable force, we must break the journey down into infinitesimally small steps, $d\mathbf{s}$. Over such a tiny step, the force is approximately constant, allowing us to calculate a tiny amount of work, $dW = \mathbf{F} \cdot d\mathbf{s}$. The total work is then the sum of all these tiny contributions, which mathematically translates to a line integral: $$W = \int \mathbf{F} \cdot d\mathbf{s}$$

### Analyzing the Setup

In our specific problem, we are given a force field defined by $\mathbf{F} = -k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}})$, where $k$ is a positive constant. Notice how the $x$-component of the force depends on the $y$-coordinate, and the $y$-component depends on the $x$-coordinate. This cross-dependence makes the force field quite interesting.

We are asked to find the work done as a particle moves from the origin $(0,0)$ to the point $(a,a)$. The problem specifies a particular path: first along the $x$-axis to $(a,0)$, and then parallel to the $y$-axis to $(a,a)$.

### The Master Equation: The Exact Differential

Let's set up our integral. The infinitesimal displacement vector in a 2D plane is $d\mathbf{s} = dx\hat{\mathbf{i}} + dy\hat{\mathbf{j}}$. Taking the dot product with our force vector gives:

$$dW = -k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}}) \cdot (dx\hat{\mathbf{i}} + dy\hat{\mathbf{j}})$$

$$dW = -k(y dx + x dy)$$

Now, we encounter a beautiful mathematical symmetry. The expression $(y dx + x dy)$ should trigger an immediate "Aha!" moment if you are familiar with the product rule in calculus. It is the **exact differential** of the product $xy$. That is, $d(xy) = y dx + x dy$.

This allows us to rewrite our work equation in a incredibly compact form:

$$dW = -k d(xy)$$

Because the integrand is an exact differential, the integral is **path-independent**. This means the force is conservative! The work done depends only on the initial and final coordinates, regardless of the specific route taken. We can integrate directly from the start to the finish:

$$W = \int_{(0,0)}^{(a,a)} -k d(xy) = -k [xy]_{(0,0)}^{(a,a)}$$

Substituting the limits, we get:

$$W = -k [ (a)(a) - (0)(0) ] = -ka^2$$

### The Alternate Route: Step-by-Step Path Analysis

Even though we know the force is conservative, it is highly instructive to calculate the work done along the specific path given in the problem to verify our result and build physical intuition.

**Path 1: From $(0,0)$ to $(a,0)$**
Along this horizontal segment, the particle is moving strictly on the $x$-axis. Therefore, the $y$-coordinate is constantly $0$, which also means $dy = 0$. Substituting $y=0$ into our force equation yields $\mathbf{F} = -kx\hat{\mathbf{j}}$.

Physically, this means the force is pointing straight down (in the negative $y$-direction), while the particle is moving horizontally (in the positive $x$-direction). Since the force and displacement are perpendicular, the dot product is zero. The force does no work on this leg of the journey.

$$W_1 = \int (-kx\hat{\mathbf{j}}) \cdot (dx\hat{\mathbf{i}}) = 0$$

**Path 2: From $(a,0)$ to $(a,a)$**
Now the particle turns and moves vertically. Along this segment, the $x$-coordinate is locked at $x=a$, meaning $dx = 0$. Substituting $x=a$ into the force equation gives $\mathbf{F} = -k(y\hat{\mathbf{i}} + a\hat{\mathbf{j}})$.

The displacement is purely vertical, $d\mathbf{s} = dy\hat{\mathbf{j}}$. When we take the dot product, the horizontal component of the force ($-ky\hat{\mathbf{i}}$) does no work because it is perpendicular to the vertical displacement. Only the vertical component ($-ka\hat{\mathbf{j}}$) contributes to the work.

$$W_2 = \int_{0}^{a} (-ka\hat{\mathbf{j}}) \cdot (dy\hat{\mathbf{j}}) = \int_{0}^{a} -ka dy$$

Since $-ka$ is a constant, the integration is straightforward:

$$W_2 = -ka [y]_{0}^{a} = -ka(a - 0) = -ka^2$$

### Final Calculation

The total work done is simply the sum of the work done on each segment:

$$W_{total} = W_1 + W_2 = 0 + (-ka^2) = -ka^2$$

Both the elegant exact differential method and the rigorous path-by-path analysis yield the exact same result, confirming the deep consistency of physics and mathematics.

Select Answer:

Visualized Solution

The Particle's Journey

The Work Done Integral

Setting up the Vectors

The Dot Product

Exact Differential

Integrating the Differential

Evaluating the Limits

Alternate Method: Path 1 Analysis

Work Done on Path 1

Alternate Method: Path 2 Analysis

Work Done on Path 2

Calculating Total Work

The Sigma Insight: Work Done by Forces

The Physics of Variable Forces

Analyzing the Setup

The Master Equation

The Alternate Route

Final Calculation

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