Work Done With/Without A Spring

In summary, the question is whether the net external work done on the system of two blocks and a spring is the same, greater than, or less than the work done in experiment 1 where two blocks are pushed together with a constant force F over a time interval deltaT. The work done in experiment 1 is W=Fx. In experiment 2, the force is still constant but there is a spring with spring constant k, so the force is now f=kx and the work done is (k/2)x^2. The presence of the spring reduces the net force on the blocks, resulting in a smaller displacement and therefore, less work done in experiment 2. This is because the spring exerts a force against the
  • #1
EzequielSeattle
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In experiment 1, two blocks of identical mass are pushed together with a force of magnitude F over a time interval deltaT. In experiment 2, the same force is applied to the same blocks over the same time, except there is a spring (initially at equilibrium) between the blocks. Is the NET EXTERNAL WORK done on the system of the two blocks and spring the same, greater than, or less than, that done in experiment one (to the system of the two blocks)?
W=Fx

It seems to me as though the work ought to be less, because the same force is applied and the spring will cause the blocks to have less displacement. Because W=Fx, and F is constant and x is getting smaller, W should get smaller. Am I right, or is the same amount of work done to the system?
 
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  • #2
EzequielSeattle said:
In experiment 1, two blocks of identical mass are pushed together with a force of magnitude F over a time interval deltaT.
Are the blocks touching? How much work is being done? What's the displacement?
 
  • #3
It does not specify in the problem. The displacement and work are not given. I need to figure out if the work in experiment 1 is greater to, equal, or less than the work in experiment 2. The only things given are a constant force F and a constant time interval deltaT.

I figure that W=Fx.W_1 = Fx_1
W_2 = Fx_2

F is constant, and because there's a spring in experiment 2, I figure that the displacement of block 2 is less (and . My intuition could also be totally wrong.
 
  • #4
Are you presenting the statement of the problem, especially the description of experiment 1, exactly as given, word for word? The phrase "pushed together" is ambiguous.
 
  • #5
The blocks start a constant distance apart in each experiment. At no point do they ever contact. In each experiment, an equal amount of force is applied to each block *toward* one another over an equal time interval. Will the *net external work* on the block-block system in experiment one be less than, equal to, or greater than, the *net external work* on the block-spring-block system in experiment two.

Hopefully that clarifies it...
 
  • #6
Saying the two blocks are pushed together with force F makes no sense unless we are to assume that there is something preventing the two blocks from moving together without any force (except some possibly arbitrarily small force to get one of the blocks moving). If there is a constant force F, then the work done would be, of course, F times the initial distance between the blocks, "Fx" as you say.

If there is a spring, with spring constant k, then the force, instead of being constant is f= kx and the work done would be [tex]\int kx dx= (k/2)x^2[/tex]. Whether this is larger than or less than Fx depends upon the relation between "F" and "K".

Your statement "x is getting smaller" doesn't really make sense. The "x" in "F= kx" is the initial distance between the two blocks, not the changing distance as they come together.
 
  • #7
EzequielSeattle said:
Hopefully that clarifies it...
I think it does, a bit. Let's further assume that the blocks are supported by a frictionless surface.

OK, now in which case will the greater displacement occur? How can you determine that?
 
  • #8
HallsofIvy said:
Saying the two blocks are pushed together with force F makes no sense unless we are to assume that there is something preventing the two blocks from moving together without any force (except some possibly arbitrarily small force to get one of the blocks moving).
Why is that?

HallsofIvy said:
If there is a constant force F, then the work done would be, of course, F times the initial distance between the blocks, "Fx" as you say.
We are told that a force F is applied, assumed constant.

HallsofIvy said:
If there is a spring, with spring constant k, then the force, instead of being constant is f= kx and the work done would be
The work done by the applied force F will still be F*displacement, regardless of the presence of the spring.
 
  • #9
Doc Al said:
I think it does, a bit. Let's further assume that the blocks are supported by a frictionless surface.

OK, now in which case will the greater displacement occur? How can you determine that?

Well, the displacement ought to be less when there's a spring because the spring will be exerting some force against the external force which you are applying, thereby acting against your force and decreasing the amount which your force is able to displace the blocks. Therefore, the work when a spring is included ought to be less. Is my logic sound?
 
  • #10
EzequielSeattle said:
Well, the displacement ought to be less when there's a spring because the spring will be exerting some force against the external force which you are applying, thereby acting against your force and decreasing the amount which your force is able to displace the blocks. Therefore, the work when a spring is included ought to be less. Is my logic sound?
Yes, very good!

The presence of the spring reduces the net force on the blocks, reducing their subsequent acceleration and displacement. Good!
 
  • #11
I appreciate your help!
 

FAQ: Work Done With/Without A Spring

What is work done with a spring?

Work done with a spring refers to the amount of energy required to compress or stretch a spring from its original position to a new position. This work is equal to the force applied to the spring multiplied by the distance the spring is compressed or stretched.

How is work done with a spring calculated?

The formula for calculating work done with a spring is W = 1/2kx^2, where W represents work, k represents the spring constant, and x represents the distance the spring is compressed or stretched.

What is the relationship between work done with a spring and potential energy?

Work done with a spring is equal to the change in potential energy of the spring. As the spring is compressed or stretched, potential energy is stored within the spring. This potential energy can be converted into work when the spring is released.

Can work be done without a spring?

Yes, work can be done without a spring. Work is defined as the force applied to an object multiplied by the distance the object moves in the direction of the force. This can be achieved without the use of a spring, such as pushing a box across the floor.

How does work done with a spring differ from work done without a spring?

The main difference between work done with a spring and work done without a spring is the presence of potential energy. When work is done with a spring, potential energy is stored within the spring and can be converted back into work. Without a spring, there is no potential energy being stored and the work is done solely by the force applied to the object.

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