Whan an elastic string of natural length 2m is fixed at one end and hangs vertically supporting a particle of mass 4Kg at the other end. It stretches to a length of 2.8m. A horizontal force of 28N is applied gradually to the mass until it is once again in equilibrium. Calculated the length and the inclination to the vertical of the string. If possible with a diagram.

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 ($x_1$):

Next, calculate the weight ($W$) of the particle:

According to Hooke's Law, the tension in the string (which is equal to the weight in this vertical setup) is $T = kx$.
So, $W = kx_1$:

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:

For horizontal equilibrium:

Divide equation $(**)$ by equation $(*)$:

The inclination to the vertical of the string is $\boxed{35.5^\circ}$.

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 $(*)$:

The new stretched length $L_2$ is the natural length plus the extension:

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