Respuesta :
From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm
The spring constant is a characteristic of a spring that measures the ratio of the force affecting the spring to the displacement caused by it. The effective spring constant is 98000 N/m.
What is Hooke's law for a spring?
According to Hooke's law, the force needed to extend or compress a spring by some distance is proportional to that distance.
Given that the cross-sectional area A of the wire is 1 mm2 and length l is 1 m. The mass m of the wire is 10 kg and stretched length l' of the wire is 1 mm.
The wire is hung from the ceiling, so gravitational acceleration is applied to the wire. The extension force of the wire is given as below.
[tex]F = mg[/tex]
[tex]F = 10 \times 9.8[/tex]
[tex]F = 98 \;\rm N[/tex]
From Hooke's law, the extension force of the metal wire is directly proportional to the extension.
[tex]F = kl'[/tex]
Where k is the effective spring constant, l' is the stretched length of the wire and F is the extension force in the wire.
[tex]98 = k \times 0.001[/tex]
[tex]k = 98000 \;\rm N/m[/tex]
Hence we can conclude that the effective spring constant is 98000 N/m.
To know more about Hooke's law, follow the link given below.
https://brainly.com/question/3355345.
