Answer:
[tex]894.12\ \text{N}[/tex]
[tex]284.013\ \text{J}[/tex]
Explanation:
U = Energy = 76 J
x = Displacement of spring = 0.17 m
k = Spring constant
Energy is given by
[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 76}{0.17^2}\\\Rightarrow k=5259.51\ \text{Nm}[/tex]
Force is given by
[tex]F=kx\\\Rightarrow F=5259.51\times 0.17\\\Rightarrow F=894.12\ \text{N}[/tex]
The magnitude of force must you apply to hold the platform is [tex]894.12\ \text{N}[/tex].
Now [tex]x=0.17+0.2=0.37\ \text{m}[/tex]
[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 5259.51\times 0.37^2\\\Rightarrow U=360.013\ \text{J}[/tex]
Additional energy
[tex]360.013-76=284.013\ \text{J}[/tex]