Solving 2 Spring Problems: Extension & Tension Calculation

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The discussion focuses on solving two problems related to the calculations of extension and tension in steel springs using Hooke's law. For the first problem, the extension of a spring with a spring constant of 25 Nm^-1 under a tension of 10N is calculated to be 0.4m, while the tension when extended by 0.50m is determined to be 12.5N. In the second problem, two identical springs are analyzed under a 40N load, leading to a tension of 20N in each spring and an extension of 100mm. The correct approach to finding the spring constant involves using the extension in the equation, resulting in 20 = 100k. The thread emphasizes the importance of correctly applying Hooke's law to derive the necessary values.
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1.A stell spring has a spring constant of 25Nm^-1.Calculate
a.the extension of the spring when the tension in it is equal to 10N,
b.The tension in the spring when it is extended by 0.50m from its unstretched length.

2.Two identical stell springs of length 250mm are suspended vertically side-by-side from a fixed point. A 40N weight is attached to the ends of the two springs. The length of each spring is then 350mm. calculate:
a. The tension in each spring,
b. The extension of each spring.
c. The spring constant of each spring.:eek:
 
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Any attempts?
 
radou said:
Any attempts?

By using Hooke's law F = kL...
a. 10 = 25 x L, L = 0.4m
b. F = 25 x 0.5 = 12.5N
 
Last edited:
Looks good.
 
radou said:
Looks good.

but i don't know how to do question 2?
a. 40/2 = 20N
b. 350mm - 250mm = 100mm
c. 20 = 350k...?
 
Last edited:
54088 said:
but i don't know how to do question 2?
a. 40/2 = 20N
Yes, of course. (Note that you were told the two springs were identical.)

b. 350mm - 250mm = 100mm
Good.

c. 20 = 350k...?
No, it is the extension that is important: 20= 100 k.
 

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