1750N/m
According to Hooke's law, the spring constant (k) of an elastic material is the ratio of the force (F) applied to the material to the extension (x) of the material caused by this force. i.e
k = F / x --------------(i)
From the question, the elastic material (spring) of 25 coils has a spring constant of 350N/m. This means that for every 350N force applied to the spring, the spring extends by 1m.
Now if the spring is cut into five equal parts each with five coils, imagine they are attached together such that the same force of 350N is applied to still cause a total extension of 1m. Each spring contributes 1/5 of this extension.
Therefore, the extension caused by each spring is 1/5 of 1m = 0.2m
Since the same force of 350N is applied, substitute F = 350N and x = 0.2m into equation (i) as follows;
k = 350 / 0.2
k = 1750N/m
Therefore, the spring constant of each of the 5-coil spring is 1750N/m
The spring constant of each of the 5-coil springs is 1,750 N/m.
The given parameters;
The total spring constant of the 25 coils is calculated as follows;
[tex]K_t = 25 \times 350 \ N/m\\\\K_t = 8,750 \ N/m[/tex]
The spring constant of each of the given 5 coils is calculated as follows;
[tex]K = \frac{8,750 }{5} \\\\K = 1,750 \ N/m[/tex]
Thus, the spring constant of each of the 5-coil springs is 1,750 N/m.
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