Question

It 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 force 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

(a)spring effective force constant =1568N/cm

(b) Yes

Step-by-step explanation:

Hooke's law is represented mathematically as, F=ke

where

• F is force applied to elastic material,

• k= spring constant

• e= extension

(a).After the maximum load is exceeded, Hooke's law doesn't apply anymore. from the problem, its stated that a maximum load of 120kg will cause an extension of 0.75cm, we will use this to determine the spring constant.

F=mg, g=9.8
`m/s^(2)`

F= 120*9.8 =1176N

From Hooke's law, k=
`(F)/(e)`

k=
`(1176)/(0.75)`

k=1568N/cm

(b). The players who stands on the scale causes a 0.48cm extension.

e= 0.48cm, k= 1568N/cm

F=Ke

F= 1568*0.48

F= 752.64N

To calculate the mass of the player we divide this force by g=9.8
`m/s^(2)`

F=mg

m=F/g

`m= ( 752.64)/(9.8) \\\\m=76.8kg`

since the rugby team is an under 85kg team, this player with 76.8kg mass is eligible