A person who weighs 716 N steps onto a spring scale in the bathroom, and the spring compresses by 0.00619 m. What is the weight of another person who compresses the spring by 0.00352 m?

Hooke's law states that the spring force is the product of a spring's constant (determines how stiff or flexible a spring is) and the displacement of the spring from equilibrium.

[tex]F_s=kx[/tex]

[tex]\hrulefill[/tex]

Solving the Problem

Reading the word problem, the forces acted upon the individual when they step on the scale are weight (downward) and spring force (upward). Since the person doesn't sink or spring into the air, we know that the net force is 0, thus the weight and spring forces are equal in magnitude.

A free-body diagram can illustrate this, a dot representing the person and arrows, one pointing up and the other down, that label the upward one "spring" and the downward one "weight".

We're told that when a weight of 716 N is applied the spring compresses by 0.00619m. Since the spring force is equal to the weight, we can equate the spring force of 716, and the displacement to x.

[tex]716=k(0.00619)[/tex]

Rearranging the equation, the spring constant is

[tex]\dfrac{716}{0.00619} =115670.44=k[/tex].

We're told that in a new scenario another person's weight compresses the spring by 0.00352. To find their weight we can take the product of the displacement and our newly found k constant.

[tex]W=F_s=(115670.44)(0.00352) \approx 407~N[/tex]