Step 1: Calculate the spring constant ($k$) of the elastic string. Given: Natural length $L_0 = 2 \text{ m}$ Stretched length $L_1 = 2.8 \text{ m}$ Mass $m = 4 \text{ kg}$ Acceleration due to gravity $g = 9.8 \text{ m/s}^2$ First, calculate the initial extension ($x_1$): $$x_1 = L_1 - L_0 = 2.8 \text{ m} - 2 \text{ m} = 0.8 \text{ m}$$ Next, calculate the weight ($W$) of the particle: $$W = mg = 4 \text{ kg} \times 9.8 \text{ m/s}^2 = 39.2 \text{ N}$$ According to Hooke's Law, the tension in the string (which is equal to the weight in this vertical setup) is $T = kx$. So, $W = kx_1$: $$k = \frac{W}{x_1} = \frac{39.2 \text{ N}}{0.8 \text{ m}} = 49 \text{ N/m}$$ Step 2: Analyze the forces when the horizontal force is applied. When the horizontal force $F_H = 28 \text{ N}$ is applied, the particle is in equilibrium under three forces: 1. The weight $W = 39.2 \text{ N}$ acting vertically downwards. 2. The horizontal force $F_H = 28 \text{ N}$ acting horizontally. 3. The tension $T$ in the string, acting along the string at an angle $\theta$ to the vertical. Let $x_2$ be the new extension of the string and $L_2$ be the new stretched length. The tension in the string is $T = kx_2 = 49x_2$. We resolve the forces into horizontal and vertical components. For vertical equilibrium: $$T \cos \theta = W$$ $$49x_2 \cos \theta = 39.2 \quad (*)$$ For horizontal equilibrium: $$T \sin \theta = F_H$$ $$49x_2 \sin \theta = 28 \quad (**)$$ Step 3: Calculate the inclination to the vertical ($\theta$). Divide equation $(*)$ by equation $()$: $$\frac{49x_2 \sin \theta}{49x_2 \cos \theta} = \frac{28}{39.2}$$ $$\tan \theta = \frac{28}{39.2}$$ $$\tan \theta = \frac{280}{392} = \frac{5}{7}$$ $$\theta = \arctan\left(\frac{5}{7}\right)$$ $$\theta \approx 35.537^\circ$$ The inclination to the vertical of the string is $\boxed{35.5^\circ}$. Step 4: Calculate the new length of the string ($L_2$). From $\tan \theta = \frac{5}{7}$, we can construct a right-angled triangle with opposite side 5 and adjacent side 7. The hypotenuse is $\sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}$. So, $\cos \theta = \frac{7}{\sqrt{74}}$. Substitute this value into equation $(*)$: $$49x_2 \left(\frac{7}{\sqrt{74}}\right) = 39.2$$ $$x_2 = \frac{39.2 \sqrt{74}}{49 \times 7}$$ $$x_2 = \frac{39.2 \sqrt{74}}{343}$$ $$x_2 \approx \frac{39.2 \times 8.6023}{343} \approx 0.9831 \text{ m}$$ The new stretched length $L_2$ is the natural length plus the extension: $$L_2 = L_0 + x_2$$ $$L_2 = 2 \text{ m} + 0.9831 \text{ m}$$ $$L_2 = 2.9831 \text{ m}$$ The length of the string is $\boxed{2.983 \text{ m}}$. Diagram: A diagram illustrating the forces and the inclination: ` Fixed point | | | Tension (T) | / |/ θ (angle to vertical) O----- F_H (Horizontal force) | | | W (Weight) | Particle `