Answer:
The magnitude of force must you apply to hold the platform in this position = 888.89 N
Explanation:
Given that :
Workdone (W) = 80.0 J
length x = 0.180 m
The equation for this work done by the spring is expressed as:
[tex]W = \frac{1}{2}k_{eq}x^2[/tex]
Making the spring constant [tex]k_{eq}[/tex] the subject of the formula; we have:
[tex]k_{eq} = \frac{2W}{x^2}[/tex]
Substituting our given values, we have:
[tex]k_{eq} = \frac{2*80}{0.180^2}[/tex]
[tex]k_{eq} = 4938.27 \ N.m^{-1}[/tex]
The magnitude of the force that must be apply to the hold platform in this position is given by the formula :
[tex]F = k_{eq}x[/tex]
[tex]F = 4938.27*0.180[/tex]
F = 888.89 N