If W1, W2 and W3 represent the work done in moving a particle from A to B along three different path

Q: Wait, why doesn't the length of the path matter?

In a conservative force field like gravity, the work done is stored as potential energy. The change in potential energy only depends on the radial distance from the mass, not the specific route taken. Any extra work done moving away is exactly canceled by the work gained moving back.

Q: What if friction was present on these paths?

If friction were present, the force would be non-conservative. Friction always opposes motion, so the work done against friction depends on the total distance traveled. In that case, the longest path would require the most work, and the equality would break.

Q: Is this path independence true for all forces?

No, this is a special property exclusive to **conservative forces** like gravity, electrostatic force, and ideal spring forces. Non-conservative forces like friction, air resistance, and viscous drag are highly dependent on the path taken.

Q: How do we mathematically prove this path independence?

For a conservative force $\vec{F}$, it can be expressed as the negative gradient of a scalar potential ($\vec{F} = -\nabla U$). Integrating this from $A$ to $B$ yields $$W = \int_A^B \vec{F} \cdot d\vec{r} = -(U_B - U_A)$$, which clearly only depends on the endpoints $A$ and $B$.

Q: Does the speed of the particle along the path change the work done?

No, the work done by the conservative force is independent of the particle's speed or kinetic energy. While the net work done equals the change in kinetic energy (Work-Energy Theorem), the work done *by gravity specifically* only depends on the spatial coordinates.

Question

If $W_1$, $W_2$ and $W_3$ represent the work done in moving a particle from $A$ to $B$ along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass $m$. Find the correct relation between $W_1$, $W_2$ and $W_3$ Options: (A) $W_1 > W_2 > W_3$ (B) $W_1 = W_2 = W_3$ (C) $W_1 < W_2 < W_3$ (D) $W_2 > W_1 > W_3$

Sigma Percentile · Accepted Answer

### The Illusion of the Path

Imagine you are standing at the base of a towering mountain. You need to reach the summit, and you have three choices. You could take the steep, direct path straight up the cliff face. You could take a long, winding trail that gently spirals around the mountain. Or, you could take a bizarre path that goes up, dips down into a valley, and then climbs back up to the peak.

Intuition might scream that the longer, more winding paths require more "work" to conquer. After all, you are walking a much greater distance!

This is the exact psychological trap presented in our problem. We have a particle moving from point $A$ to point $B$ along three distinct paths: a curved upper path (Path 1), a direct straight line (Path 2), and a curved lower path (Path 3). We are asked to compare the work done, $W_1$, $W_2$, and $W_3$, along these routes.

### The Invisible Hand of Gravity

To solve this, we must look at the environment. The problem explicitly states that the particle is moving in the **gravitational field** of a point mass $m$.

This is the master key to the entire puzzle. Gravitational force is not just any force; it belongs to an elite category of forces known as **conservative forces**.

Other members of this exclusive club include electrostatic forces between charges and the restoring force of an ideal spring. But what makes a force "conservative"?

### The Perfect Energy Accounting System

When a force is conservative, it means the universe acts as a perfect accountant. Any work done against this force is not lost to the void as heat, friction, or sound. Instead, it is meticulously stored in the system as **potential energy**.

Mathematically, the work done by a conservative force is defined purely by the negative change in potential energy between the final and initial states. We can write this elegantly as:

$$W = -\Delta U = -(U_{	ext{final}} - U_{	ext{initial}})$$

Notice what is missing from this equation? There is no variable for distance traveled. There is no variable for the shape of the curve.

### The Grand Conclusion: Path Independence

Because the work done depends *only* on the potential energy at the starting point and the ending point, we arrive at one of the most beautiful principles in physics: **Path Independence**.

The work done by a conservative force is completely independent of the path taken. It simply does not matter how you get from point $A$ to point $B$.

Let's apply this profound truth to our particle. 
- For Path 1, the journey starts at $A$ and ends at $B$.
- For Path 2, the journey starts at $A$ and ends at $B$.
- For Path 3, the journey starts at $A$ and ends at $B$.

Since the initial position ($A$) and the final position ($B$) are identical for all three paths, the change in potential energy $\Delta U$ is exactly the same for all of them.

Therefore, the work done must be perfectly equal across all routes. The universe doesn't care about the detours; it only cares about the destination.

$$W_1 = W_2 = W_3$$

This elegant realization leads us directly to our final answer. The correct relation is equality across all paths.

Select Answer:

Visualized Solution

The Setup

The Force Field

Nature of Gravity

Conservative Forces

Path Independence

Applying the Concept

Final Conclusion

The Sigma Insight: Work Done by Forces

The Illusion of the Path

The Invisible Hand of Gravity

The Perfect Energy Accounting System

The Grand Conclusion

Similar Questions

Consider a force $F = - x \hat{i} + y \hat{j}$ . The work done by this force in moving a particle from point $A (1, 0)$ to $B (0, 1)$ along the line segment is (all quantities are in SI units)

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force $F = (α y \hat{i} + 2 α x \hat{j}) N$ , where x and y are in meter and $α = - 1 N/m^{- 1}$ . The work done on the particle by this force $F$ will be ____ Joule.

A force $F = (5 \hat{i} + 3 \hat{j} + 2 \hat{k}) N$ is applied over a particle which displaces it from its origin to the point $r = (2 \hat{i} - \hat{j}) m$ . The work done on the particle in joule is

A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled 3 m is

A time dependent force $F = 6 t$ acts on a particle of mass $1 kg$ . If the particle starts from rest, the work done by the force during the first $1 s$ will be

A block of mass $m$ is kept on a platform which starts from rest with constant acceleration $\frac{g}{2}$ upwards as shown in figure. Work done by normal reaction on block in time $t$ is

A force acts on a 2 kg object, so that its position is given as a function of time as $x = 3 t^{2} + 5$ . What is the work done by this force in first 5 seconds?

A body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 g h$ . The value of workdone by the air-friction is