Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m