Explanation:
The given data is as follows.
Work done by the force, (W) = 79.0 J
Compression in length (x) = 0.190 m
So, formula for parallel combination of springs equivalent is as follows.
[tex]k_{eq} = K_{1} + K_{2}[/tex]
= 2k
Hence, work done is as follows.
W = [tex]\frac{1}{2}k_{eq} \times x^{2}[/tex]
[tex]k_{eq} = \frac{2W}{x^{2}}[/tex]
= [tex]\frac{2 \times 79}{(0.19)^{2}}[/tex]
= [tex]\frac{158}{0.0361}[/tex]
= 4376.73 N/m
Hence, magnitude of force required to hold the platform is as follows.
F = [tex]k_{eq}x[/tex]
= [tex]4376.73 N/m \times 0.19 m[/tex]
= 831.58 N
Thus, we can conclude that magnitude of force you must apply to hold the platform in this position is 831.58 N.
