How Much Work Is Needed to Stretch Two Springs and Move a Cube?

[duplaimp]

Homework Statement

Two identical ideal massless springs have unstretched lengths of 0.25m and spring constants of 200N/m. The springs are attached to a small cube and stretched to a length L of 0.32m. One spring on the left and one on the right. An external force P pulls the cube a distance of D = 0.020m to the right and holds it there. What is the necessary work to put the strings and the cube like in figure A?

Homework Equations

W = [itex]\frac{1}{2}[/itex]k([itex]xi^{2}-xf^{2}[/itex])

The Attempt at a Solution

I tried with the above equation with xi = 0.25 and xf = 0.32 but the work value isn't right (it should be 0.98J). Any hint to solve this?

 

[barryj]

I don't know where the formula comes from but you might look at the stored energy before the block moves and the stored energy after the block moves. This might be the work?

Are you sure of the answer of 0.98J?

Do you mean the work to put the block like in figure B, not A?

 

[haruspex]

I tried with the above equation with xi = 0.25 and xf = 0.32

That would be the work required to get to position A starting from a position in which both springs are relaxed. It wouldn't involve B at all.
I have a strong feeling you have paraphrased the original question. It is not clear what is being asked. Please quote the exact text.

Edit: That's wrong. I should have written: "That would be the work required to stretch the springs from an initial extension of 0.25 to an extension of 0.32. The 0.25 is an unstretched length."
I agree with TSny's interpretation, on the proviso that there are more parts to the question, involving set-up B.

 

[TSny]

I believe the question is asking for the work required to set up the system in the initial state of figure A. That is, what's the work necessary to stretch both springs from their natural length to 0.32 m?

 

[duplaimp]

I believe the question is asking for the work required to set up the system in the initial state of figure A. That is, what's the work necessary to stretch both springs from their natural length to 0.32 m?

Yes, it is asking what is the work requires to set up it like in figure A. I thought it was that and used the equation above but the result is wrong

 

[barryj]

I disagree. If the question is just about setting up configuration A, then why is there any mention of.."An external force P pulls the cube a distance of D = 0.020m to the right and holds it there." and why does the diagram B show the mass displaced to the right? The answer is still in question, IMHO.

 

[duplaimp]

I disagree. If the question is just about setting up configuration A, then why is there any mention of.."An external force P pulls the cube a distance of D = 0.020m to the right and holds it there." and why does the diagram B show the mass displaced to the right? The answer is still in question, IMHO.

Some questions in this exercise set I am solving have extra uneeded information. The answer is just what is the work to setup the system like in figure A

 

[haruspex]

W = [itex]\frac{1}{2}[/itex]k([itex]xi^{2}-xf^{2}[/itex])

I tried with the above equation with xi = 0.25 and xf = 0.32

Did you see my edit at post #3? In the equation, x_i is an initial extension. In the problem, 0.25 is a relaxed length, not an extension.

 

[duplaimp]

Did you see my edit at post #3? In the equation, x_i is an initial extension. In the problem, 0.25 is a relaxed length, not an extension.

So what would be xi and xf?

 

[haruspex]

So what would be xi and xf?

When a spring is at its relaxed length, by how much is it being extended?

 

[duplaimp]

When a spring is at its relaxed length, by how much is it being extended?

0

But it says "The springs are attached to a small cube and stretched to a length L of 0.32m" so that's why I tried with xf and xi. So should I try with

K = [itex]\frac{1}{2}[/itex]k(x[itex]^{2}[/itex]-0)?

This then becomes

x = 0.32-0.25 = 0.07

K = [itex]\frac{1}{2}[/itex]200(0.07[itex]^{2}[/itex]) = 0.49J * 2 (two springs) = 0.98J

Is this correct?

 

[haruspex]

K = [itex]\frac{1}{2}[/itex]200(0.07[itex]^{2}[/itex]) = 0.49J * 2 (two springs) = 0.98J

Is this correct?

Yes, that gives the work done in creating the set-up in A, which is what the question seems to be asking. And you stated that 0.98J is known to be the correct answer.