Question

is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective spring constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85 kg team?

Answer 1

Answer:

a) 156960 N/m

b) Yes the player can play on the team.

Explanation:

F = Force

a = Acceleration due to gravity = 9.81 m/s²

m = Mass

k = Spring constant

x = Deformation length of spring

F = ma

From Hooke's law

$F=kx\\\Rightarrow ma=kx\\\Rightarrow k=\frac{ma}{x}\\\Rightarrow k=\frac{120\times 9.81}{0.0075}\\\Rightarrow k=156960\ N/m$

The effective spring constant is 156960 N/m

When x = 0.48 cm = 0.0048 m

$m=\frac{kx}{a}\\\Rightarrow m=\frac{156960\times 0.0048}{9.81}\\\Rightarrow m=76.8\ kg$

The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.