Using calculus to solve a physics question

this is so frustrating.. can someone please help??

5) Suppose that 2j of work is needed to stretch a spring from its natural lenght of 20cm to a length of 30cm. How much work is needed to stretch it from a length of 22cm to 28cm?
Use hookes law

The book's answer is 1.2j ... i got 4.99 j.. wtf

Here is what i did:

so f(x)=kx

so 2j=k (0.1)

k = 20N

and then i integrated and did the steps.. but i dont know what i did wrong

Hooke's Law is an equation for the force.

What you want is the energy.

This is where the calculus comes into it: $F(x) = kx = \frac{\text{d}V}{\text{d}x}$, where $V$ is the potential energy.

Integrate both sides w.r.t. $x$, then substitute to get the value of $k$. Then you can find the answer to the question.

Also, note that $x$ is the displacement from the spring's natural length.

For what it's worth, I agree with the book's answer.

thanks for the reply.. whats w.r.t?

Sorry, it's short for 'with respect to'.

So: $\displaystyle V=\int F \text{d}x = \int kx \text{ d}x$

hmmm so v= Fx= kx^2/2 ?? and then i do F= w/d so 2j / 0.10m so i get 20 N and then i substitute the 20 in the x? I am feeling kinda lost aye

You have the right expression for the energy, $W=\frac{1}{2}kx^2$, but the wrong value for $k$.

Given $W=2$ when $x=(0.30 - 0.20) = 0.1$, substitute those values in again and see what you get. Don't forget to square the $x$!

I dont understand why we are using the equation W =0.5k x^2 .. why cant we use w = F d

anyways i got k=400?

far out i am so confused

If you take a force of the form $F=c$, where $c$ is some constant, integrating will give you $W=cx$, which is equivalent to $W=Fd$.

$W=Fd$ is for situations where the force does not change with distance, $d$. In the case of a spring, the force grows larger as $x$, the displacement from its natural length, increases.

Anyway, now that you have the right value of $k$, you can move onto finding what the question asked for.

Remembering that the natural length of the spring is 20cm = 0.2m, the potential energy associated with an extension from 22cm to 28cm will be simply:

$W=\frac{1}{2}k(0.28 - 0.20)^2 - \frac{1}{2}k(0.22 - 0.20)^2$

That is, the difference between the energy stored in the spring at extensions of 28cm and 22cm.

Post again if this still sounds like a mystery to you.

how come u didnt use 0.30 in ur equation and umm okay i kinda get it but not really.. my book explains this a lot different

Maybe it'd be better to check with your teacher or someone, then ; I don't want to start teaching you an alternative method that'll only serve to confuse you.

Calculus can be a bit confusing at first, but practise makes perfect.

I'm off to have some food now, but I should be back later.

EDIT: Who negged you?

its probably ma teacher anyways i kinda dont get this...and i have my exam in a couple of hours...i guess im gonna **** up this question...i'll try my best though and go over what u said

Good luck with your exam!

Dont see why calculus was necessary? Couldn't you just use E=1/2ke^2 for both equations?

Yes, you do that equation for both. But calculus is necessary to get to it, as far as I'm aware.

I was under the impression, given the way the OP worded the question, that the equation for the potential energy stored in a spring was not 'assumed prior knowledge'.