Question

A person who weighs 846 N steps onto a spring scale in the bathroom, and the spring compresses by 0.574 cm. (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by 0.314 cm?

Answer 1

Step-by-step explanation:

It is given that,

Weight of the person, W = F = 846 N

When the person steps onto a spring scale in the bathroom, the spring compresses by 0.574 cm, x = 0.00574 m

(a) The force acting on the spring is is given by Hooke's law as :

`F=-kx`

`k=(F)/(x)`

`k=(846\ N)/(0.00574\ m)`

k = 147386.7 N/m

(b) If the spring is compressed by 0.314 cm or 0.00314 m, weight of the person is given by again Hooke's law as :

`F=kx`

`F=147386.7\ N/m* 0.00314\ m`

F = 462.7 N

Hence, this is the required solution.