Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of

Q: Wait, why don't we need to integrate the force along the elliptical path?

Because the applied force is constant in both magnitude and direction. For any constant force, the work done is simply $\vec{F} \cdot \Delta\vec{r}$, regardless of the actual path taken between the initial and final points.

Q: Why is the work done by gravity negative?

Gravity acts downwards, but the block is moving upwards from $y = 0$ to $y = 4\text{ m}$. Since the displacement is opposite to the direction of the force, the work done is negative.

Q: Does the normal force do any work?

No. The normal force is always perpendicular to the surface of the rail, and therefore perpendicular to the instantaneous velocity of the block. The dot product of perpendicular vectors is zero, so $W_N = 0$.

Q: How did we find the net displacement to be $5\text{ m}$?

The block moves from point $P(3,0)$ to point $Q(0,4)$. The net displacement is the straight line connecting these two points. Using the distance formula or Pythagorean theorem, $\sqrt{3^2 + 4^2} = 5\text{ m}$.

Q: What if the force was always tangential to the elliptical path?

If the force was always tangential, its direction would be constantly changing. In that case, it would NOT be a constant force vector, and you would have to perform a line integral $\int \vec{F} \cdot d\vec{s}$ along the elliptical arc to find the work done.

Question

Consider an elliptically shaped rail $PQ$ in the vertical plane with $OP = 3	ext{ m}$ and $OQ = 4	ext{ m}$. A block of mass $1	ext{ kg}$ is pulled along the rail from $P$ to $Q$ with a force of $18	ext{ N}$, which is always parallel to line $PQ$ (see figure). Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n 	imes 10)	ext{ J}$. The value of $n$ is (take acceleration due to gravity $= 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### The Setup: A Deceptive Path

Imagine you are standing in front of a giant elliptical rail. A block of mass $1	ext{ kg}$ is resting at point $P(3,0)$ and needs to be pulled to point $Q(0,4)$. The path is curved, and your first instinct might be to set up a complex line integral to calculate the work done.

But wait, look closely at the applied force! The problem states that the force of $18	ext{ N}$ is **always parallel to the line $PQ$**.

### The Masterstroke: Path Independence

This single phrase is the key to unlocking the entire problem. Because the force has a constant magnitude ($18	ext{ N}$) and a constant direction (parallel to $\vec{PQ}$), it is a **constant force vector**.

One of the most beautiful properties of physics is that the work done by a constant force is completely independent of the path taken. It doesn't matter if the block moves along an ellipse, a zigzag, or a spiral staircase. The work done depends *only* on the net displacement vector $\vec{s}$.

$$ \vec{s} = \vec{r}_Q - \vec{r}_P = \vec{PQ} $$

The length of this displacement vector is simply the hypotenuse of a right triangle with sides $3	ext{ m}$ and $4	ext{ m}$.

$$ |\vec{s}| = \sqrt{3^2 + 4^2} = 5	ext{ m} $$

### The Arsenal: Work-Energy Theorem

Now we bring in our heavy artillery: The **Work-Energy Theorem**. It states that the net work done by all forces acting on an object equals its change in kinetic energy.

$$ W_{	ext{net}} = \Delta K = K_f - K_i $$

Let's break down the work done by each individual force:

**1. The Applied Force ($W_F$):**
Since the applied force $\vec{F}$ is perfectly parallel to the displacement vector $\vec{s}$, the angle between them is $0^\circ$.

$$ W_F = |\vec{F}| |\vec{s}| \cos(0^\circ) = 18 	imes 5 = 90	ext{ J} $$

**2. Gravity ($W_{mg}$):**
Gravity is a conservative force acting downwards. The block moves upwards by a vertical distance of $h = 4	ext{ m}$. Because the displacement is opposite to the force, the work done is negative.

$$ W_{mg} = -mgh = -(1)(10)(4) = -40	ext{ J} $$

**3. The Normal Force ($W_N$):**
At every instant, the normal force from the rail is perpendicular to the block's velocity. The dot product of perpendicular vectors is zero, so the normal force does absolutely no work.

$$ W_N = 0	ext{ J} $$

### Final Calculation

Assuming the block is pulled from rest, its initial kinetic energy $K_i$ is zero. We can now sum the work done and find the final kinetic energy $K_f$:

$$ W_F + W_{mg} + W_N = K_f $$

$$ 90 - 40 + 0 = K_f $$

$$ K_f = 50	ext{ J} $$

The problem tells us that the final kinetic energy is given by the expression $(n 	imes 10)	ext{ J}$. Equating our result to this expression:

$$ 50 = n 	imes 10 $$

$$ n = 5 $$

And just like that, by recognizing the path independence of a constant force, we bypassed a terrifying integral and arrived at an elegant, clean solution!

Enter Numerical Value:

Visualized Solution

The Physical Setup

Forces in Action

Constant Force Vector

Path Independence

Net Displacement $s$

Calculating $W_{F}$

Calculating $W_{m g}$

Calculating $W_{N}$

Work-Energy Theorem

Finding $K_{f}$

Comparing to find $n$

The Sigma Insight: Work Done by Forces

The Setup

The Masterstroke

The Arsenal

Final Calculation

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