To solve this problem it is necessary to apply the equations given in Newton's second law as well as Hooke's Law.
Since Newton's second law we have that force is
F = mg
Where,
m = mass
g = gravity
From the hook law, let us know that
F = -kx
Where
k = Spring constant
x = Displacement
Re-arrange to find k,
[tex]k = -\frac{F}{x}[/tex]
PART A ) We can replace the Newton definitions here, then
[tex]k = -\frac{mg}{x}[/tex]
Replacing with our values we have that
[tex]k = \frac{120*9.8}{-0.75*10^{-2}}[/tex]
[tex]k = 1.57*10^5N/m[/tex]
Therefore the required value of the spring constant is [tex]1.57*10^5N/m[/tex]
PART B) We can also equating both equation to find the mass, then
[tex]mg = -kx[/tex]
[tex]m = \frac{-kx}{g}[/tex]
Replacing with tour values we have
[tex]m = \frac{1.568*10^5(-9.48*10^{-2})}{9.8}[/tex]
[tex]m = 76.8Kg[/tex]
Therefore the mass of the player can be of 76.8Kg, then the player is eligible to play because the mass is less than 85Kg
