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## Homework Statement

When a spring fixed at one end is pulled by a force of 8N, the extension is 40mm. Two such springs are joined in series and is pulled to produce a total extension of 40mm. What is the strain energy in the springs.

## Homework Equations

strain energy, U = Fe / 2 or

U = F

^{2}/2k

*e is the extension while k is the spring constant.

( I found this equation online, correct me if this is wrong. )

## The Attempt at a Solution

I calculate the strain energy of 1 spring first, which is

8 x 0.04m / 2 = 0.16J

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for one spring, the spring constant should be

F = ke

8 = 0.04k

k = 200

For two spring, the spring constant is

1/k = 1/k

_{1}+ 1/k

_{2}

1/k = 1/100

k = 100

hence, the force used to made an extension of 0.04m is

F = 100 x 0.04

= 4N

Thus, the strain energy is

U = F

^{2}/ 2k

= 16 / 200

= 0.08J

or

U = Fe /2

= 4 x 0.04 / 2

= 0.08J

The answer provided by the book is 0.32J, which is differ with the answer I calculated, either one string or two strings. Can anyone correct the mistakes I made while doing calculation? Or the answer given is incorrect?