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,

$k = \frac{F}{x}$

$k = \frac{685}{0.88*10^-2}$

$k = 77840.9N/m$

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.