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

Step 1: Calculate the displacement $PQ$ and the vertical height.

The points $O$, $P$, and $Q$ form a right-angled triangle.

$$PQ = \sqrt{OP^2 + OQ^2}$$ $$PQ = \sqrt{3^2 + 4^2}$$ $$PQ = 5\text{ m}$$

The vertical height $h$ that the block rises is equal to $OQ$.

$$h = 4\text{ m}$$

Step 2: Calculate the work done by the applied force.

Since the force is constant in magnitude and direction, the work done is independent of the path.

$$W_F = F \cdot PQ$$ $$W_F = 18 \times 5$$ $$W_F = 90\text{ J}$$

Step 3: Calculate the work done by gravity.

The gravitational force acts downwards while the block rises vertically.

$$W_g = -mgh$$ $$W_g = -1 \times 10 \times 4$$ $$W_g = -40\text{ J}$$

Step 4: Apply the work-energy theorem to find the kinetic energy at $Q$.

The change in kinetic energy is equal to the net work done on the block.

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

Step 5: Find the value of $n$.

Equate the calculated kinetic energy to the given expression.

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

Answer: The value of $n$ is 5.