Question

A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the spring constant? N/m (b) What is the weight of another person who compresses the spring by 0.38 cm?

Answer 1

To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,

`F = kx`

Where,

k = Spring constant

x = Displacement

Initially our values are given,

`F = 685N`

`x = 0.88 cm`

PART A ) With this values we can calculate the spring constant rearranging the previous equation,

`k = (F)/(x)`

`k = (685)/(0.88*10^-2)`

`k = 77840.9N/m`

PART B) Since the constant is unique to the spring, we can now calculate the force through the new elongation (0.38cm), that is

`F = kx`

`F = (77840.9)(0.0038)`

`F = 295.79N`

Therefore the weight of another person is 265.79N.