Answer:
The ideal spring is 0.030 meters above the starting position.
Explanation:
Let consider that spring is modelled by the Hooke's Law, which is represented by the following expression:
[tex]F = k\cdot \Delta x[/tex]
Where:
[tex]F[/tex] - Force exerted on the ideal spring, measured in newtons.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]\Delta x[/tex] - Spring elongation, measured in meters.
The elongation is now cleared:
[tex]\Delta x = \frac{F}{k}[/tex]
The initial force experimented by the spring is due to the weight of bananas and potatoes.
[tex]F = (m_{b} + m_{p})\cdot g[/tex]
Where:
[tex]m_{b}[/tex], [tex]m_{p}[/tex] - Masses of bananas and potatoes, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If [tex]m_{b} = 1\,kg[/tex], [tex]m_{p} = 2\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]k = 655\,\frac{N}{m}[/tex], the initial elongation of the ideal spring is:
[tex]F = (1\,kg + 2\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 29.421\,N[/tex]
[tex]\Delta x_{o} = \frac{29.421\,N}{655\,\frac{N}{m} }[/tex]
[tex]\Delta x_{o} = 0.045\,m[/tex]
The final elongation is obtained after eliminating the influence of potatoes due to gravity. That is to say:
[tex]F = m_{b}\cdot g[/tex]
[tex]F = (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 9.807\,N[/tex]
The final elongation of the ideal spring is:
[tex]\Delta x_{f} = \frac{9.807\,N}{655\,\frac{N}{m} }[/tex]
[tex]\Delta x_{f} = 0.015\,m[/tex]
The displacement of the spring due to the removal of potatoes is:
[tex]d = \Delta x_{o} - \Delta x_{f}[/tex]
[tex]d = 0.045\,m-0.015\,m[/tex]
[tex]d = 0.030\,m[/tex]
The ideal spring is 0.030 meters above the starting position.
