To solve the equation we proceed to use the law of conservation of the material, so
For A)
[tex]mgh = \frac{1}{2}k(h-L)^2[/tex]
[tex]75*9.8*h = \frac{92.5}{2}(h-35)^2[/tex]
Solving for h,
[tex]h= 67.8m[/tex]
For B)
First obtain x,
[tex]mg=kx[/tex]
[tex]x= \frac{mg}{k} = \frac{75*9.8}{92.5} = 7.95m[/tex]
Then,
[tex]mg(L+x) - \frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]
[tex](75)(9.8)(42.95)-\frac{1}{2}(92.5)(7.95)^2 = \frac{1}{2}(75)v^2[/tex]
[tex]v= 27.6 m/s[/tex]
For 3)
Greater than. Because the final distance is greather than 35mm.
