June 1st, 2014, 02:28 PM  #1 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4  Hooke's Law Confusion
If 3 J of work are needed to stretch a spring from 8 cm to 10 cm and 5 J are needed to stretch it from 10 cm to 12 cm, what is the natural length of the spring? So above is the problem I have been wrestling with for the past hour. I know how to find Work if I am given a force such as 40N. to stretch the spring from it's natural length of 100cm to say 105cm. I do know then how to measure the distance from 100cm to 105cm = 5 cm which in turn = .05meters. and then how to maneuver from f(x) = k(x) and so, f(.05) = 40 and so... k = 800 and then how to integrate our f(x) = 800x and to follow through with the integration process to obtain Work = X joules. What I cannot seem to figure out on my own is (and for the question I posted here) we are not given a force and are not given in the spring's natural length. We are given the amount of work for "two" seperate scenarios and so I am trying to set up the equation for force and am hoping if I can solve for force that I can find when the work done = 0 (or signifies the spring is at it's natural resting place (length). Sadly, I have been unsuccessful trying to comprehend this problem and thus I am posting it here on the forum. *So, is my thinking correct trying to solve for the unknown variable (force) ...plugging in for WORK and how would you advance from here? I have looked over similar Hooke's Law problems but they all seem to give the Force and Natural Length of the Spring and I have noticed that a lot of these Calculus 2 Homework problems in WebAssign have been a variation that I feel tries to get the student to "think outside of the box" , etc. Thanks Forum. 
June 1st, 2014, 07:23 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra 
From memory, Hooke's Law says that $F = kx$ where x is the extension and k is a constant that tells you how easy it is to stretch the spring. And total length $l = n + x$ where n is the natural length. So, you have two unknowns (n and k) and two equations, so we ought to be able to get values for booth. Can you set up the two integrals, using n and k and the information given in the question? 
June 1st, 2014, 10:44 PM  #3 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4 
I will try

June 1st, 2014, 11:05 PM  #4 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4 
actually v8 I am not sure that I understand how to set up the Integral with the two equations and two unknowns. I need help

June 2nd, 2014, 07:53 AM  #5 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4 
v8 I think that F= kx leaves me with 2 unknowns in this equation alone. I know to picture a number line ( or assign a # line ) to the spring. Thus, when we move backwards (or compress the spring) we are moving in a negative direction (hence the k) Thus if we have a 10 cm's then  (10)(x) gives a positive answer of simply kx. Now, problem I was thinking is F = kx I do not have k nor F (force) so I am thinking there is 2 unknowns in this equation and 1 unknown in the equation you gave of length l = natural length n + extension x. The equation has given us the value for Work for each equation. i.e., one is 3J and for the other it is 5j. Both move the same distance (the first from 8cm to 10cm ...thus a distance of (.02)meters and the other 10cm to 12 cm. Thus also a distance of (.02)meters. Although now I recongnize what you have made mention of total distance which I can clearly see as (.04)meters. So, F, n and k I think I don't have. Last edited by mr. jenkins; June 2nd, 2014 at 07:57 AM. 
June 2nd, 2014, 03:44 PM  #6  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications 
Hi mr. jenkins, Quote:
 
June 2nd, 2014, 07:36 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra 
Yes, you either have $F$ or you have $kx$, you don't have both, so $F$ and $k$ are only a single unknown in the system. Or perhaps more accurately, if you know $k$, you can find $F$ and vice versa. You are looking to create integrals that look like $$\int_{a}^{b}{Fx}dx$$ where the limits need to be the extension, not the total length of the spring. Try to create them, and post your efforts. We'll see if you need further assistance. 
June 3rd, 2014, 11:57 AM  #8 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4 
Well as far as setting up the Integrals, I am still stuck with the following thinking: F = kx Ok, since W = Force*Displacement I see the following and please correct me where I am wrong. The question says 3 Joules are needed to stretch a spring from 8 cm to 10 cm then it says:and 5 Joules are needed to stretch spring from 10 cm to 12 cm Thus, The first informations I would think I could set up like this: 3 = Fd 3 = F(2 cm) and for every problem in the unit thus far they have had us to change the cm to meters so... 3 = F(.02) Now, since Work = Force * Displacement I have 3 = F(.02) Thus, Force = 150 and on the other equation that looks about the same 5 = Fd = Force = 250 Now, I need to know is this correct or incorrect? and if b and a are the limits of the extension (meaning I think not the compression but opposoite) we would still have .02 for each one so what? b =.02 and a =0 ? Now if f(x) = kx and I did the first part right, I now know that k(.02) = 150 and k(.02) =250 giving us k values of 7,500 and 12,500. Thus Integrals would look like f(x) = k(x) = 150 ... k = 7500 ... INT from a to b of 7500x dx would go (7500/2)x^2 = 3,750(.02)  3750(0) = 75 That would be on the first Integral and second would just be 6,250(.02)6,250(0) = 125 and if this is at all correct I am thinking I should have taken final integral value of larger subtract the smaller which would give 12575 = 50 Ok guys 
June 3rd, 2014, 02:56 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra 
W = Fd is fine for a constant force. Ours isn't constant, because as we displace the object, the spring stretches more, so the force increases. Our integral will look like $$W = \int_a^b Fx dx$$ We've decided that, for a spring, F=kx where $x$ is the extension. It is also the displacement against the resisting force, the one against which we are doing work. So $$W = \int_a^b kx^2 dx$$ We are given two sets of conditions, the Work Donw to strecth the spring between two total lengths (NOT extensions). We can use these conditions to produce a pair of integrals on the model above. Can you fill in the blanks yet? 
June 3rd, 2014, 03:35 PM  #10 
Senior Member Joined: Dec 2013 Posts: 137 Thanks: 4 
no i don't really understand


Tags 
confusion, hooke, law 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Orientation on Calculus Work Problem. Hooke's Law.  Dan3500  Calculus  2  August 5th, 2013 05:21 PM 
Plz help me get rid of my confusion !  saravananr  Algebra  3  February 24th, 2013 04:20 AM 
Confusion.  CherryPi  Algebra  14  August 15th, 2011 05:00 AM 
Help confusion!  flower555  Calculus  1  October 16th, 2010 04:38 PM 
Using Hooke's Law  linlinrocks  Calculus  1  May 9th, 2009 04:49 AM 