Spring compressed 1m by 10N force: where does the other half the work go?

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alcatras
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HI! I have a problem here about work and energy. !Help is needed!

"If a spring (on horizontal surface) is pressed down by 10N force the max. compress is 1m. What is the : a)Work done b)Potential energy gained by spring?"

First of all I found 10=k x 1 k=10N/m

Then work done W=F.x=10.1=10C

Ep=kX^2/2=10x1/2=5C

As a result Energy gained is half of the work done.Is it correct? Where is the other half of work? :smile: :smile:

THX! THX! THX!
 

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alcatras said:
First of all I found 10=k x 1 k=10N/m

Then work done W=F.x=10.1=10C
To find the work done in compressing the spring, think about starting from equilibrium and applying a force that is just enough to compress the spring, gradually building up to 10 N.

Ep=kX^2/2=10x1/2=5C

As a result Energy gained is half of the work done.Is it correct? Where is the other half of work?
If you could figure out how to do this, you would be rich. Unfortunately, energy is conserved and work cannot be less than the potential energy.

AM
 
alcatras said:
HI! I have a problem here about work and energy. !Help is needed!

"If a spring (on horizontal surface) is pressed down by 10N force the max. compress is 1m. What is the : a)Work done b)Potential energy gained by spring?"

First of all I found 10=k x 1 k=10N/m
Yes, F= kx. F= 10N, x= 1m so k= F/x= 10 N/m

Then work done W=F.x=10.1=10C

NO! "W= Fx" is only true if F is a constant. I don't know at what level you are doing this but what you need to do is integrate Fdx over the range of motion: [tex]W= \int_0^1(10x)dx= 5x^2\|_0^1= 5[/tex] N-m (Joules).

Ep=kX^2/2=10x1/2=5C

Yes! Exactly what I just said. Where do you think that formula came from?

As a result Energy gained is half of the work done.Is it correct? Where is the other half of work? :smile: :smile:

THX! THX! THX!

"Energy gained" is exactly equal to the "work done".
 
There are two possibilities:

1. The force wasn't always ten Newtons, it was just enough overcome the spring force and get it compressed by 1 m (eg, if someone pushed the spring down with their hand). In this case, the force varies with distance and you would need to integrate over it to get the work done, which would come out to be 5 J (what is a C?)

2. The force really is always ten Newtons (eg, a 1 kg object under gravity), and the spring starts uncompressed with the mass is at rest. Then while the compression is less than 1 m, the mass on the spring is accelerating because there is a net force on it. Eventually it will reach a point where the force of the spring equals ten Newtons, but it will still have a velocity. It will then go some extra distance until all this kinetic energy is stored in the spring (after which it would bounce back up and go into simple harmonic motion). If this is what they mean by max compression, the value of k will be greater than ten, and again, energy conservation will work out.
 
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Eeh, it could also have been meant that the student should understand that half the work from the constant, applied force was lost (from the pool of mechanical energy), for example through heating up the spring.

If this were the case, I think the exercise is dumb at the outset.
It does not seem probable to me at least that the mechanical energy loss through heating is comparable to the gain in potential energy.
 
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