# "Work" Related problems

#### shamieh

##### Active member
&quot;Work&quot; 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?

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

#### Bacterius

##### Well-known member
MHB Math Helper
Re: &quot;Work&quot; 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?

- - - 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.