Question

A weight of 12.0 N hangs motionless from a spring, with a spring constant of 10.0 N/m. How far is the spring stretched from its original length?

0.120 m
0.220 m
0.833 m
1.20 m

Answer 1

0.120 m

Step-by-step explanation:

F=kx where F is force and k is constant, x is extension from its original length

Making x the subject then substituting 12 N for F and 10.0 N/m for k then

`x=\frac {F}{k}=\frac {12 N}{10 N/m}=0.12 m`

Answer 2

1.2m

Step-by-step explanation:

According to hookes law which States that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically, F = ke where;

F is the applied force = 12.0N

k is the elastic constant = 10N/m

e is the extension = ?

Substituting the values into the formula to get the extension, we have e = F/k

e ,= 12N/10N/m

e = 1.2m

This shows that the string have been stretched by 1.2m from its original length