# Homework Help: Work Done With/Without A Spring

1. Oct 17, 2014

### EzequielSeattle

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?

Last edited by a moderator: Oct 17, 2014
2. Oct 17, 2014

### Staff: Mentor

Are the blocks touching? How much work is being done? What's the displacement?

3. Oct 18, 2014

### EzequielSeattle

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. Oct 18, 2014

### Staff: Mentor

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. Oct 18, 2014

### EzequielSeattle

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. Oct 18, 2014

### HallsofIvy

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 $$\int kx dx= (k/2)x^2$$. 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. Oct 18, 2014

### Staff: Mentor

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. Oct 18, 2014

### Staff: Mentor

Why is that?

We are told that a force F is applied, assumed constant.

The work done by the applied force F will still be F*displacement, regardless of the presence of the spring.

9. Oct 18, 2014

### EzequielSeattle

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. Oct 18, 2014

### Staff: Mentor

Yes, very good!

The presence of the spring reduces the net force on the blocks, reducing their subsequent acceleration and displacement. Good!

11. Oct 18, 2014