A particle acted on by constant forces 4i + j - 3k and 3i + j - k is displaced from the point i + 2j

Q: Why can't I just calculate the work done by each force separately and add them up?

You absolutely can! The beauty of physics is that it is consistent. The work done by the resultant force is mathematically identical to the sum of the work done by individual forces because the dot product is a linear operator. However, calculating the resultant force first is a 'pro-tip'—it saves you time and reduces the chance of arithmetic errors. It is about working smarter, not harder.

Q: Does the path taken by the particle matter here?

Great question! In this specific problem, the forces are constant. When forces are constant, the work done is path-independent. It only depends on the initial and final positions. This is a hallmark of conservative fields, though here it is simply a property of constant vectors. Whether the particle moved in a straight line or a zig-zag, the work done remains the same.

Q: What if the dot product gives a negative value? Does that mean the work is wrong?

Not at all! A negative dot product simply means the angle between the force and the displacement is obtuse (greater than 90 degrees). Physically, this means the force is acting in a direction that opposes the motion, effectively trying to slow the particle down. It is a perfectly valid physical result, not an error.

Question

A particle acted on by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$ is displaced from the point $\hat{i} + 2\hat{j} - 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$. The total work done by the forces is

Options:

(A) 50 units

(B) 20 units

(C) 30 units

(D) 40 units

Sigma Percentile · Accepted Answer

### The Dance of Vectors: A Journey Through Work and Space

Imagine, for a moment, that you are standing in a vast, empty three-dimensional void. In front of you, a tiny particle is suspended, waiting for a push.

This isn't just any particle; it is a test subject for the laws of mechanics. We are going to observe how this particle responds to the invisible hands of force. This problem is not just about plugging numbers into a formula; it is about understanding the geometry of energy.

### Phase 1: The Principle of Superposition

Our particle is initially at position $\vec{r}_A = \hat{i} + 2\hat{j} + 3\hat{k}$. Suddenly, two forces, $\vec{F}_1 = 4\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{F}_2 = 3\hat{i} + \hat{j} - \hat{k}$, begin to pull on it.

You might feel the urge to calculate the work done by $\vec{F}_1$ and then the work done by $\vec{F}_2$ separately. While that is mathematically sound, it is like walking around the block to get to the house next door. Physics offers us a shortcut: the **Principle of Superposition**.

We can replace these two distinct forces with a single, equivalent force, $\vec{F}_{net}$. Think of this as the 'total push' the particle feels. By adding the vectors component-wise, we simplify our universe:

$$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (4+3)\hat{i} + (1+1)\hat{j} + (-3-1)\hat{k} = 7\hat{i} + 2\hat{j} - 4\hat{k}$$

This single vector, $7\hat{i} + 2\hat{j} - 4\hat{k}$, is the heartbeat of our problem. It represents the combined intent of both forces acting in unison.

### Phase 2: The Displacement Vector

Now, the particle moves. It travels from its initial home at $\vec{r}_A$ to a new destination, $\vec{r}_B = 5\hat{i} + 4\hat{j} + \hat{k}$.

In the world of mechanics, we don't care about the path taken—the curves, the loops, or the detours. We only care about the displacement, $\vec{d}$. The displacement is the straight-line arrow that connects the start to the finish.

Calculating this is a simple subtraction of vectors:

$$\vec{d} = \vec{r}_B - \vec{r}_A$$

$$\vec{d} = (5-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} = 4\hat{i} + 2\hat{j} - 2\hat{k}$$

Look at this vector: $4\hat{i} + 2\hat{j} - 2\hat{k}$. It tells us exactly how the particle shifted in space. It moved $4$ units in the $x$-direction, $2$ in the $y$-direction, and retreated $2$ units in the $z$-direction.

### Phase 3: The Dot Product—The Soul of Work

We have our net force, and we have our displacement. Now, we must calculate the work done.

Work is not just force times distance; it is the projection of force onto the direction of motion. This is why we use the dot product. The dot product, $\vec{F} \cdot \vec{d}$, measures how much of the force is actually 'helping' the particle move in the direction of the displacement.

When we write $W = \vec{F}_{net} \cdot \vec{d}$, we are essentially asking: "How much of this force is aligned with this movement?" Substituting our values, we get:

$$W = (7\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k})$$

Remember the golden rule of dot products: $\hat{i} \cdot \hat{i} = 1$, but $\hat{i} \cdot \hat{j} = 0$. The cross-terms vanish into thin air, leaving us with only the products of the corresponding components. This is the elegance of the dot product—it filters out the noise and leaves us with the essence.

### Phase 4: The Final Calculation

Let us perform the arithmetic with precision:

$$W = (7 	imes 4) + (2 	imes 2) + (-4 	imes -2)$$

$$W = 28 + 4 + 8$$

$$W = 40 	ext{ units}$$

Twenty-eight plus four is thirty-two. Adding eight brings us to forty. The total work done is exactly **$40 	ext{ units}$**.

### Conclusion: The Elegance of Physics

Take a moment to appreciate what you have just done. You took a complex scenario involving multiple forces and a 3D displacement, and you distilled it into a single, clean number.

You didn't get lost in the complexity; you used the tools of vector algebra to slice through it. This is the essence of JEE Advanced physics. It is not about memorizing formulas; it is about visualizing the physical reality and applying the right mathematical tool to reveal the truth.

Select Answer:

Visualized Solution

Initial Position of the Particle

Forces Acting on the Particle

Net Force (Superposition Principle)

Calculating the Resultant Force $F$

Final Position of the Particle

Displacement Vector $d$

Calculating Displacement $d$

Work Done by a Constant Force

Setting up the Dot Product

Executing the Dot Product

Final Calculation

Summary and Conclusion

The Sigma Insight: Scalar (Dot) Product

The Dance of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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