Potential energy stored in a spring

[stfrankies240]

man you have helped me a lot..thank u very much

[owaisoz]

is it that when force varies, we use integration to get work ?

[keagleeye]

elastic potential energy is actually 1/2KdeltaX^2 ...

[mrRnf11]

@keagleeye It means the same thing, final-initial extension is assumed.

[shikharsingh007]

W=f.d and my physics= xd

[aznf0rl1fe]

why does khan academy need money? isnt it nonprofit lol and only sal is doing it.

[LookItsZander]

@aznf0rl1fe If you gave all these lessons, you would spend alot of time doing these videos. which means you can't have a job.which means you have no money to live. and so, ta-da!

[ZacharySmith89]

@aznf0rl1fe False, KhanAcademy now has many people behind the scenes.

[lumaix]

I'd been watching khans videos all day long and feel asleep at the beginning of this video after pausing it... I dreamt a dream of people running around shouting their velocities and acceleration and woke up to lots of people outside running in the street shouting at each other- except it wasn't velocities O_O

[toiya666]

thanx a lot pal..I feel like staying at home rather than going to college,its worth,the time n everything when ppl like you exist..

[Believia]

Yes, complex math is needed. Good job.

The mgh approach can be used if in a hurry, with a good precision, but only if we get a number of different known weights and a metric line, empirically.

[Believia]

Here is my logique;

Tell me if I'm wrong;

- i have 75 kg, i go & sit on a spring, it deforms 10 cm while i loose 10 cm of altitude, so i lost 75*9.82*0.1 = 73J of Penergy.

- block me and the spring somehow and remove it from under me.

You got 73 J stored in it.

True ?

[Believia]

Not true, 'cause the spring can't elevate me back by 10 cm. Maybe half, 36.5 J.

But, after i "armed" the spring, some kid , which is half ofmy wght [37.5 kg] sit on the spring and release it slowly... than the kid will pe pulled up by 9.91 cm ???

[0.0991*9.82*37.5 = 36.5 J]

Dt tell me that with the kid half my wght, a spring "armed" by me, will regain 99% of the compression created by my weight.

Lookslike spring ctt is ncssary and the maths.

But let's make it without them, pls someone help.

[Believia]

@Believia . I lost some decimals. Ok, it is 36.5 J.  My previous message has a very stupid part, sorry.

[sponsoredwalk1]

The reason he is using ½kx² comes from the description of work as;

∫Fdx = ∫kxdx = ½kx² where the force on a spring is proportional to the distance it is pulled mediated by a spring constant k.

Also, using Newtons 3rd law, if a force is used to compress a spring a distance x, the force exerts an opposite force on whatever is compressing it, and there is energy stored in the spring with the "potential" to bring the spring back to it's normal position.

[sodr2]

F = k x

W = F d

Then shouldn't work = k xd = kx^2?

Why do you have to divide by 2, or calculate the area *under* the slope??

[onlyAerik]

@sodr2: because energy is the derivative of work, we must integrate a graph of work to find the potential energy.

If F(x) = kx^n, then integral of F(x) = (k/(n+1))x^(n+1) where n=1, givng us (k/(1+1))x^(1+1) = (k/2)x^2

[moonpocket234]

This helped me so much! thank you :)

[polos505]

I thought that the

kenitic energy =1/2*mv(v)

potential energy=mgh

how come you used potential energy=1/2kx(x) ? anyway,thanks for the beautiful explaining.I learn a lot..

[krirosfor]

potential energy is mgh, BUT in a spring it's 1/2 kx (X)

[coolboy03k]

yes, right

[scourgeofutopia]

mgh is gravitational potential energy rather than 1/2k(x^2) which is variable force potential energy

[ureasonnable]

the law of conservation of energy states that energy can not be lost; which mean that : work=potential energy = kinetic energy. that mean that mgh=1/2*mv^2 .

[JBilal]

@polos505 It's been a year since you posted this so I doubt you need clarification but Gravitational Potential Energy is entirely different from Elastic Potential Energy, which explains why they are two entirely different equations.

[newhyperbeamattack]

@polos505

Compressing a spring is different from lifting a mass in the air. If you lift a mass in air from a distance of y=0 to y=20, you are exerting the same force at all times when you do work against gravity. However, when you compress a spring, the amount of force you need to apply is different at different x values (for example, if you want to compress from x=0 to x=5, you need to apply more force when you are at x=4 than at x=3). This is how I understood it and this makes them different.

[idricool]

Thanks, you're great ! I've learned so much from you.