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WorkEnergy Theorum: Spring potential energy vs Kinetic Energy 
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#1
Oct1009, 07:22 PM

P: 104

1. The problem statement, all variables and given/known data
A 1350kg car rolling on a horizontal surface has a speed v = 40 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.5 m. What is the spring constant of the spring? Ignore Friction and assume spring is massless. 2. Relevant equations [tex] W = \Delta E[/tex] [tex] E_{pspring} = \frac{1}{2}(kx^2) [/tex] [tex] E_k = \frac{1}{2}(mv^2) [/tex] 3. The attempt at a solution First right off the bat, i converted 40 km/h to its m/s equivalent of aprox. 11.11 m/s i state the law of conservation of energy: Energy before = Energy after Therefore: [tex] E_k = E_{pspring} \frac{1}{2}(mv^2) = \frac{1}{2}(kx^2) [/tex] then i isolate k [tex] k = \frac{mv^2}{x^2} [/tex] now heres the issue, is x negative? because the displacement is against the direction of motion? and 2.5m = x, (2.5)^2 gives me a answer of 4266 Nm but (2.5)^2 is entirely different.. This has been a long lasting math issue for me. And what if x is positive? i know k MUST be positive right? 


#2
Oct1009, 08:21 PM

HW Helper
P: 4,433

(2.5)^2 is correct. There is no negative energy in the nature.



#3
Oct1009, 08:25 PM

HW Helper
P: 3,394

There is no minus sign in mv^2 = kx^2
or in k = mv^2/x^2. No way you can get k negative! The minus sign in F = kx is supposed to help keep track of the fact that the force of the spring is opposite to the direction of stretch but it does seem to have a habit of getting in the way. k is ALWAYS positive. 


#4
Oct1109, 08:41 AM

P: 104

WorkEnergy Theorum: Spring potential energy vs Kinetic Energy
Thanks, the negative sign on mv^2 was an algebra error... Thanks for the clarification guys!



#5
May1810, 09:56 AM

P: 69

Your attempt is correct but you missed somethingthat for a spring if you take natural length as the datum, the force on change in length is given as:
[tex] \vec{F}= k \vec{x} [/tex] and hence work done by a spring against external forces [tex] W_{s}=\int\vec{F}\vec{.dx} [/tex] over the required limits in our case the answer is [tex]W_{s}=\frac{kx^{2}}{2}[/tex] as [tex]W_{s}=\Delta E[/tex] [tex]\Delta E=\frac{mv^{2}}{2}[/tex] the change part was where you lost it all...the KE FELL TO ZERO. HENCE A NEGATIVE CHANGE. 


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