
#1
Apr509, 03:27 AM

P: 38

I have gotten the sign wrong on every single spring forcerelated problem. I'm pretty sure I'm working out the problems correctly, but the answer turns out to be negative when I have it positive, or positive when I have it negative. Of course, on other occasions, having the wrong sign throws off the rest of the problem. Once I switch the sign, the problem works out correctly (according to the answers posted in the back of the book), but I have no real understanding of why I'm doing it.
I am using the equations given to me in the book, namely F = kx and W = 1/2kx^2, and I'm not leaving out the negative sign when I work the problems. I think I keep in mind that the force is supposed to be opposite the displacement. So what do I keep doing wrong here? Can someone please give me some practical or conceptual advice to start getting these problems right? 



#2
Apr509, 03:46 AM

Sci Advisor
HW Helper
P: 4,301

There are so many places you can get a sign wrong. As you say, it is important to keep in mind that the force act opposite to the displacement (i.e. it wants to restore the spring to equilibrium).
I think it is hard to say something specific, perhaps it helps if you post a typical problem with your wrong answer so we can see where it is going wrong. 



#3
Apr509, 04:27 AM

P: 4,513

Keep in mind that there are two forces. The force of the spring on some object and the force of the object on the the spring. Secondly, a positive force along X is in the positive X direction.
[tex]F=kx[/tex] Which one is this? Is it the force on the spring or by the spring? On and By are your two key words. Always think of them in bold font and you should get your signs correct. Edit: there is another idea to keep in mind dealing with vector quantities that can be confusing at first. So if the above is not enough, and you are using force as a vector, there can be one more thing to keep in mind. 



#4
Apr509, 05:58 PM

P: 38

work done by spring forces  positive or negative
^ If there are two forces... then which is the force for which I'm supposed to solve?
Well, here's the latest problem I just solved. I'm not sure it's a good example of what I've been doing wrong, but I did get the signs wrong, so we'll see: In the arrangement of Figure 711 (which is basically a standard block attached to spring resting horizontally), we gradually pull the block from x = 0 to x = +3.0 cm, where it is stationary. Figure 736 gives the work that our force does on the block. We then pull the block out to x = +5.0 cm and release it from rest. How much work does the spring do on the block when the block moves from xi = +5.0 cm to (a) x = +4.0 cm, (b) x = –2.0 cm, and (c) x = –5.0 cm? So I found the spring constant k: W = 1/2kx^2 0.9 = 1/2k(0.03)^2 k = 2000 N/m Now to solve part a: W = 1/2kxf^2  1/2kxi^2 W = 1/2(2000)(0.05^2)  1/2(2000)(0.04^2) W = 2.5 + 1.6 = 0.9 J (The correct answer is 0.9 J.) Now to solve part b: W = 1/2kxf^2  1/2kxi^2 W = 1/2(2000)(0.05^2)  1/2(2000)(0.02^2) W = 2.5 + 0.4 = 2.1 J W = 2.1 J (The correct answer is 2.1 J.) Part c was 0 because the net work from x=0.05 cm to x=0.05 cm is 0. Thanks in advance. 



#5
Apr609, 02:29 AM

Sci Advisor
HW Helper
P: 4,301

I think the problem is in the calculation of your work.
The potential energy of the spring is [itex]U = \frac12 k x^2[/itex] from which you can determine the spring constant. Quote Wikipedia: "(Note that potential energy of a spring is always nonnegative.)". The work you are doing to stretch the spring is getting stored in the spring as potential energy. I.e. you are doing work on the block so if you let W be the work done by the spring on the block then U = W (you are adding energy to the spring while W is like energy released by the spring in the process). Does that make sense? 


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