Question

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?

Answer 1

Answer:

0.126m

Explanation:

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