The horizontal force applied to the block is approximately 1,420.84 N
The known parameters;
The mass of the block, w₁ = 400 kg
The orientation of the surface on which the block rest, w₁ = Horizontal
The mass of the block placed on top of the 400 kg block, w₂ = 100 kg
The length of the string to which the block w₂ is attached, l = 6 m
The coefficient of friction between the surface, μ = 0.25
The state of the system of blocks and applied force = Equilibrium
Strategy;
Calculate the forces acting on the blocks and string
The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N
The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N
Let T represent the tension in the string
The upward force from the string = T × sin(θ)
sin(θ) = √(6² - 5²)/6
Therefore;
The upward force from the string = T×√(6² - 5²)/6
The frictional force = (W₂ - The upward force from the string) × μ
The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25
The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)
∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]
[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]
The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N
Therefore;
The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68
The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]
The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P
Where;
P = The horizontal force applied to the block
P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84
The horizontal force applied to the block, P ≈ 1,420.84 N
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