Question

an elastic spring has a length L1 when tension in it is 4N it length is L2 when tension in it is 5N what will be its length when tension in its 9N

Answer 1



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When the tension on a wire is 4N its length is l1. When the tension on the wire is 5N its length is l2. Find its natural length

one year ago

Answers : (2)

let x1 and x2 be the extension with 4N and 5N respectively.

then the eqns. would be:

4 = k * x1  and  5  = k * x2

and let l be the natural length of the wire.

Hence,

x1 + l = l1

and

x2 + l = l2 

put x1 and x2 values in Tensions after dividing them respectively:

Therefore,

4/5 = x1/x2 = > (l1 – l)/(l2 – l)

4l2 – 4l = 5l1 – 5l

l = 5l1 – 4l2

hence would be the nautral length.

Answer 2

Answer:

Explanation:

The force exerted by a spring is given by F = kx  

Where k is the spring constant  and x is the elongation in the spring from its natural length, L. Thus,  

= 4 = k(a-L)

= 5 = k(b-L)

Solving both the equations -  

= 4/(a-L) = 5/(b-L)

= 4b-4L = 5a-5L

= L = 5a-4b

Thus,  

k = 4/(a-5a+4b) = 4/(4b-4a)

k = 1/(b-a)

Using F = kx with F = 9, where let the new length be = d.

9 = k(d-L)

9 = (d-5a+4b)/(b-a)

9b-9a = d-5a+4b

d = 5b-4a.

Thus, if 9N is applied the length will be 5b-4a.