Why is the 1/2 randomly added to the spring work equation?

[shamieh]

"Work" Related problems

I'm sure this is a simple answer..Was just curious though why are we multiplying \(\displaystyle \frac{1}{2}\) by everything?

Here is the problem.
if 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14cm, what is the natural length of the spring?

so I said \(\displaystyle \int^{0.12}_{0.10} k*x dx = 6 J\)

which is \(\displaystyle k * \frac{x^2}{2} | 0.10 to 0.12 = 6 J\)

so why are they doing it like this in the book --> \(\displaystyle \frac{1}{2}k * x^2 | 0.10 to 0 .12 = 6j\)

where is the 1/2 randomly coming from?(Worried)

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Wait are they just multiplying through to get rid of fractions?

 

[Nono713]

Re: "Work" Related problems

I'm sure this is a simple answer..Was just curious though why are we multiplying \(\displaystyle \frac{1}{2}\) by everything?

Here is the problem.
if 6 J of work is needed to stretch a spring from 10 cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14cm, what is the natural length of the spring?

so I said \(\displaystyle \int^{0.12}_{0.10} k*x dx = 6 J\)

which is \(\displaystyle k * \frac{x^2}{2} | 0.10 to 0.12 = 6 J\)

so why are they doing it like this in the book --> \(\displaystyle \frac{1}{2}k * x^2 | 0.10 to 0 .12 = 6j\)

where is the 1/2 randomly coming from?(Worried)

- - - Updated - - -

Wait are they just multiplying through to get rid of fractions?

$$\frac{1}{2} x^2 = \frac{x^2}{2}$$

I suppose they are putting the $\frac{1}{2}$ separately to make it clearer where it is coming from (i.e. integrating a linear force over a distance). But the two results are the same.