an elastic spring has a lenghtl(1) when it is stretched with a force of 2N and a lenght l (2) when it is stretched with a force of 3N. What will be the length of the spring if it is stretched with a force of 5N

Dear Student,
Please find below the solution to the asked query:

Let the natural length of the spring be 'l'. Then,
The force on a spring is given by 

F = k(spring constant) × displacement due to force

and, 
2 = k × (l1-l)
or, 
2 = kl1 - kl
​kl = kl​1-2     ---- (1)

3 = k ×​ (l3-l)
3 = kl​2-kl
kl = kl​2-3   --- (2)
comparing 1 and 2 
 kl​1-2  = kl​2-3
k (l2-l1) = 1
k = 1l2-l1     ---- (3)

substituing th evalue of 'k' in eq 1 we have,

2 = 1l2-l1×l1-l2l2-l1 = l1-ll = l1 - 2l2-l1l = 3l1-2l2      ------ (4)
Now, 

if x' is the displacement for the force of 5N
then,

5 = k × x'
5 = 1l2-l1×x'x' = 5×l2-l1

If the total length of the spring after application of 5N force is L
then,

L = l + x'
L = 3l1-2l2 + 5l2  - 5l1
L = 3l2 - 2l1


Hope this information will clear your doubts about Laws of motion.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Satyendra singh

 

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