Solving the Spring Work & Energy Problem - Help Needed!

In summary: I mean, it would have to be a pretty inefficient spring, or the force would have to be applied for a very long time to generate heat to the tune of 5 joules, no?In summary, the conversation discusses a problem regarding work and energy, specifically related to a spring being compressed by a 10N force and the resulting work done and potential energy gained. The formula for work done is provided and it is explained that the energy gained is equal to the work done. There are two possible explanations for the other half of work done, either the force varied or the spring was not fully compressed. It is recommended to check out a thread for further understanding on the topic.
  • #1
alcatras
3
0
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!
 
Physics news on Phys.org
  • #2
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
 
  • #3
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".
 
  • #4
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.
 
Last edited:
  • #6
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.
 
Last edited:

1. How do I determine the spring constant for a given spring?

The spring constant, also known as the force constant, can be calculated by dividing the force applied to the spring by the displacement of the spring from its equilibrium position. This can be done through experiments or by using the equation k = F/x, where k is the spring constant, F is the force applied, and x is the displacement.

2. How do I calculate the potential energy of a spring?

The potential energy stored in a spring can be calculated using the equation U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This equation is derived from Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.

3. How do I find the work done by a spring?

The work done by a spring can be calculated using the equation W = 1/2kx^2, where W is the work done, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This equation is derived from the definition of work, which is the product of force and displacement in the direction of the force.

4. Can I apply the equations for springs to other systems?

Yes, the equations for springs can be applied to other systems that exhibit similar behaviors, such as elastic bands, rubber bands, and bungee cords. These systems also follow Hooke's Law and have a spring constant that can be calculated in the same way.

5. How does the mass of the object affect the spring's behavior?

The mass of the object attached to the spring does not affect the spring's behavior. The spring constant and the potential energy stored in the spring will remain the same regardless of the mass attached to it. However, the displacement and work done by the spring will vary depending on the mass of the object.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
461
  • Introductory Physics Homework Help
Replies
13
Views
258
  • Introductory Physics Homework Help
Replies
29
Views
907
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
728
  • Introductory Physics Homework Help
Replies
3
Views
360
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
441
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top