Question

the length of a spring is l1andl2 when stretched with a force if 4N and 5N respectively lts natural lenth is

Answer 1

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−l)2(l2−l1) = (l1−l)l = l1 − 2(l2−l1)l = 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.
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smart Abhishek Yadav ......