**Given:**

* Force acting on the block, F = 800 N
* Inclination of the plane, θ = 30 degrees
* Friction coefficient of the plane, μ = 0.3
* Length of the displacement, d = 5 m

**To find:**

The minimum work done needed to pull the block up to a length of 5m.

**Solution:**

To find the minimum work done, we need to consider the forces acting on the block and the displacement of the block.

**1. Weight of the block:**

The weight of the block acts vertically downward and can be calculated using the formula:

W = m * g

Where m is the mass of the block and g is the acceleration due to gravity.

Given that the weight of the block is 800 N, we can calculate the mass using the formula:

m = W / g

Taking the value of g as 9.8 m/s^2, we get:

m = 800 / 9.8 = 81.63 kg

**2. Normal force:**

The normal force acts perpendicular to the plane and can be calculated using the formula:

N = m * g * cos(θ)

Where θ is the angle of inclination.

Substituting the values, we get:

N = 81.63 * 9.8 * cos(30) = 706.9 N

**3. Frictional force:**

The frictional force acts parallel to the plane and can be calculated using the formula:

f = μ * N

Substituting the values, we get:

f = 0.3 * 706.9 = 212.07 N

**4. Net force:**

The net force acting on the block can be calculated by considering the horizontal and vertical components of the weight and applying trigonometry.

The horizontal component of the weight is given by:

F_horizontal = m * g * sin(θ)

Substituting the values, we get:

F_horizontal = 81.63 * 9.8 * sin(30) = 400 N

The net force can be calculated by subtracting the frictional force from the horizontal component of the weight:

Net force = F_horizontal - f

Net force = 400 - 212.07 = 187.93 N

**5. Work done:**

The work done can be calculated using the formula:

Work = force * displacement * cos(θ)

Substituting the values, we get:

Work = 187.93 * 5 * cos(30) = 1627.18 J

Therefore, the minimum work done needed to pull the block up to a length of 5m is 1627.18 J.