Question

15. A spring has a constant of 152 when a force of 50.92 N is applied to it. How far did the spring stretch?

O=0.321 m
O z=0.418 m
O z=0.289 m
Oz=0.335 m
O None of these is correct.

Answer 1

Oz=0.335 m

Step-by-step explanation:

Given parameters;

Spring constant = 152 N/m

Force applied = 50.92N

Unknown:

How far did the spring stretch = ?

Solution:

For springs;

F = ke

F is the force

K is the spring constant

e is the extension

Now insert the parameters and find e;

50.92 = 152 x e

e = 0.335m