A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the distance the spring is stretched from its unstretched length?

0.126m

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            (i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               (ii)

But;

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

From equation (ii), it implies that;

F = W = m x g

Now substitute F = m x g into equation(i) as follows;

F = k x e

m x g = k x e      (iii)

From the question;

m = m1 = 3.5kg

k = 278N/m

Substitute these values into equation (iii) as follows;

3.5 x 10 = 278 x e

35 = 278e

Now solve for e;

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

double the distance = divide strength by 4

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)