Kinetic Energy of Block Pulled Along Elliptical Rail

To find the kinetic energy of the block upon reaching $Q$, we can apply the Work-Energy Theorem, which states that the net work done on an object is equal to the change in its kinetic energy:

$W_{\text{net}} = \Delta K = K_Q - K_P$

Given that the block starts from rest at point $P$, its initial kinetic energy is $K_P = 0$. Therefore, the kinetic energy at $Q$ is:

$K_Q = W_F + W_g$

where $W_F$ is the work done by the applied pulling force and $W_g$ is the work done by the gravitational force.

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Step 1: Work Done by the Applied Force ($W_F$)

Let the origin $O$ be at $(0, 0)$. Then, the coordinates of points $P$ and $Q$ are:

• $P = (3\text{ m}, 0\text{ m})$
• $Q = (0\text{ m}, 4\text{ m})$

The displacement vector $\vec{r}_{PQ}$ from $P$ to $Q$ is: $\vec{r}_{PQ} = (0 - 3)\hat{i} + (4 - 0)\hat{j} = -3\hat{i} + 4\hat{j}$

The magnitude of the displacement along the straight line $PQ$ is: $|\vec{r}_{PQ}| = \sqrt{(-3)^2 + 4^2} = 5\text{ m}$

Since the applied force $\vec{F}$ has a constant magnitude of $18\text{ N}$ and remains constantly parallel to the line $PQ$, $\vec{F}$ is a constant vector pointing in the direction of $\vec{r}_{PQ}$. The work done by a constant force depends only on the total displacement vector, regardless of the path taken along the elliptical rail:

$W_F = \vec{F} \cdot \vec{r}_{PQ} = F \cdot |\vec{r}_{PQ}| = 18\text{ N} \times 5\text{ m} = 90\text{ J}$

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Step 2: Work Done by Gravity ($W_g$)

The force of gravity acts vertically downwards ($\vec{F}_g = -mg\hat{j}$). The change in vertical height as the block moves from $P$ ($y_P = 0\text{ m}$) to $Q$ ($y_Q = 4\text{ m}$) is $\Delta y = 4\text{ m}$.

$W_g = -mg \Delta y = -(1\text{ kg}) \times (10\text{ ms}^{-2}) \times (4\text{ m}) = -40\text{ J}$

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Step 3: Kinetic Energy at $Q$

Summing the work done by both forces:

$K_Q = W_F + W_g = 90\text{ J} - 40\text{ J} = 50\text{ J}$

We are given that $K_Q = (n \times 10)\text{ Joules}$:

$n \times 10 = 50 \implies n = 5$