Step-by-step Solution

1. Understand the Setup and Identify Forces

   • The system is a block of mass m = 1 kg.
   • The block moves on a vertical elliptical rail from point P to point Q.
   • The coordinates of the points are P = (0, 3 m) and Q = (4 m, 0), taking O as the origin.
   • The forces acting on the block are:
     • Gravitational force, $F_g = mg$, acting downwards.
     • An applied force, F = 18 N, which is always parallel to the line PQ.
     • The normal force from the rail, N, which is always perpendicular to the path. Since it's perpendicular to the displacement, it does no work.
   • We assume the block starts from rest, so its initial kinetic energy $KE_P = 0$.
   • We need to find the final kinetic energy $KE_Q$.

2. Apply the Work-Energy Theorem The work-energy theorem states that the net work done on an object equals its change in kinetic energy. $W_{net} = \Delta KE = KE_Q - KE_P$ The net work is the sum of the work done by all forces: $W_{net} = W_F + W_g + W_N$ Since $W_N = 0$ and $KE_P = 0$: $KE_Q = W_F + W_g$

3. Calculate the Work Done by Gravity (W_g) The work done by the conservative gravitational force is the negative change in gravitational potential energy. $W_g = -\Delta U_g = -(U_{g,Q} - U_{g,P})$ Let the potential energy be zero at the ground level (y=0). The initial height at P is $h_P = 3$ m and the final height at Q is $h_Q = 0$ m. $U_{g,P} = mgh_P = (1 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m}) = 30 \text{ J}$ $U_{g,Q} = mgh_Q = (1 \text{ kg})(10 \text{ m/s}^2)(0 \text{ m}) = 0 \text{ J}$ Therefore, the work done by gravity is: $W_g = -(0 - 30) \text{ J} = 30 \text{ J}$ Alternatively, since the block's vertical displacement is 3 m downwards, gravity does positive work: $W_g = mg\Delta h = (1)(10)(3) = 30$ J.

4. Calculate the Work Done by the Applied Force (W_F) The applied force F has a constant magnitude of 18 N and is always parallel to the line PQ. This means the force vector $\vec{F}$ is constant in both magnitude and direction throughout the motion.

   • The displacement vector from P to Q is $\vec{d} = \vec{r}_Q - \vec{r}_P = (4\hat{i} + 0\hat{j}) - (0\hat{i} + 3\hat{j}) = 4\hat{i} - 3\hat{j}$.
   • The magnitude of the displacement vector is $|\vec{d}| = |\vec{PQ}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m.
   • The force $\vec{F}$ is in the same direction as the displacement vector $\vec{d}$. The work done by a constant force is given by $W_F = \vec{F} \cdot \vec{d}$. Since $\vec{F}$ is parallel to $\vec{d}$, the dot product simplifies to the product of their magnitudes. $W_F = F \times |\vec{d}|$ $W_F = 18 \text{ N} \times 5 \text{ m} = 90 \text{ J}$ It is important to note that for a constant force, the work done is independent of the path taken between the initial and final points.

5. Calculate the Final Kinetic Energy (KE_Q) Now, we substitute the calculated work values into the work-energy theorem equation: $KE_Q = W_F + W_g = 90 \text{ J} + 30 \text{ J} = 120 \text{ J}$

6. Determine the value of n The problem states that the final kinetic energy is $KE_Q = (n \times 10)$ Joules. $120 = n \times 10$ $n = \frac{120}{10} = 12$

Conclusion and Comparison with Stored Answer

The derived value for n is 12. The stored correct answer is 5. There is a disagreement.

Let's analyze the discrepancy. The solution derived above is based on a straightforward application of the work-energy theorem. The calculations for work done by gravity (30 J) and the applied force (90 J) are robust based on the problem statement.

If we assume the stored answer n=5 is correct, the final kinetic energy would be $KE_Q = 5 \times 10 = 50$ J. Using the work-energy theorem, $KE_Q = W_F + W_g$, we would have: $50 J = W_F + 30 J$ This implies that the work done by the applied force should be $W_F = 20$ J.

For $W_F$ to be 20 J, given that the force acts over a displacement whose component along the force is 5 m, the magnitude of the force would need to be: $F = W_F / d = 20 J / 5 m = 4$ N.

This suggests that the question likely contains a typo, and the applied force should have been 4 N instead of 18 N. However, solving the problem as stated gives n=12.