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Force of Spring and Potential Energy of Spring Pushing a Box 
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#1
May2912, 05:23 AM

P: 32

1. The problem statement, all variables and given/known data
A horizontal spring with spring constant 97.9 N/m is compressed 18.2 cm and used to launch a 2.96 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.156. How far does the box slide across the rough surface before stopping? 2. Relevant equations Usp = spring potential = 1/2 (k)Δx^2 F = kΔx Work = FΔx 3. The attempt at a solution Hi, so I'm not really looking for the answer... I got the answer oddly, there is just this little weird thing I'm dealing with. Ok so I found the potential energy from the spring using the equation above and got 1.6 J now because the block is sent flying the potential from the spring is converted to kinetic, which should be the same amount I believe... Now I was wondering why when I find the Force of the spring which comes to be 17.8178 N and multiply by the compression which is 0.182m, I don't get the same number as the potential energy? The work is done on... the block which has no kinetic energy, being at rest soooo the work should be equal to the change in kinetic energy which should be +1.6 J, but when I multiply 17.8178 with the compression distance 0.182m I get 3.24 J which is about 2x 1.6 which is a coincidence? or I am missing something. Anyway yeah this seems like something that should be fine but I'm dazzled by the different numbers I'm getting, any clarification would be awesome 


#2
May2912, 05:33 AM

P: 854

Hi RadiantL!
The work done by the spring varies with x, so you cannot use the simple work formula, W=F.d, as this gives you work for a constant force. You would have to integrate to find the actual work, whose magnitude, in fact would come out equal to the potential energy of the spring. That's why you do not get an exact answer from F.d, but instead get double the value, because force is linearly varying, which you can verify from the integral that would give you a 1/2 term. Hope I didn't confuse you further. 


#3
May2912, 05:39 AM

P: 963

The force in not constant, proportional to distant.
If you use integral, the ∫f(x)dx=1/2kx^{2} 


#4
May2912, 05:40 AM

P: 32

Force of Spring and Potential Energy of Spring Pushing a Box
Ah I understand now! thank you very much, both of your replies cleared things up pretty nicely :P It is very much appreciated!



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