Question

When a force of 36 Newtons is applied to springs S1 and S2, the displacement of the springs is 6 centimeters and 9 cm, respectively.

What is the difference between the spring constants of the two springs?

Answer 1

Answer:
`200 (N)/(m)`

Explanation:

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force
`F` applied to it, as long as the spring is not permanently deformed:

`F=k \Delta x`

Where:

`F=36 N`

`k` is the elastic constant of the spring

`\Delta x` is the displacement of the spring after applying the force

In this case we have two springs with constants
`k_(1)` and
`k_(2)`, displacement
`\Delta x_(1)=6 cm (1 m)/(100 cm)=0.06 m` and
`\Delta x_(2)=9 cm (1 m)/(100 cm)=0.09 m`, and the same force is applied to both.

For spring 1:

`k_(1)=(F)/(\Delta x_(1))`

`k_(1)=(36 N)/(0.06 m)`

`k_(1)=600(N)/(m)`

For spring 2:

`k_(2)=(F)/(\Delta x_(2))`

`k_(2)=(36 N)/(0.09 m)`

`k_(2)=400(N)/(m)`

Calculating the difference between them:

`k_(1)-k_(2)=600(N)/(m)-400(N)/(m)`

Finally:

`k_(1)-k_(2)=200(N)/(m)`