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Spring Constant problem 
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#1
Dec106, 03:44 AM

P: 263

Well.. I've been understanding spring constants up until this point. I just did a lab on them and understood its effects in parallel and series of springs, and I've understood it in regards to a mass and a coil in which the mass is just hanging from a coil.. but now that there is a tabletop and distance the block travels in the air, it's confusing a bit.. so here is the question, equations I think are important, and my work. I was hoping if someone could help me understand this better by telling me where I went wrong and any helpful tips. I really did well on my last test because people (docAl and olderdan, and a few others ^^) were really patient with me and helped me understand it rather than giving me the answer. So I think it's best if I'm not given the answer but only guidelines. Thanks ^^
1. The problem statement, all variables and given/known data One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance X. The block is released, and strikes the floor at a horizontal distance D from the edge of the table. Air resistance is negligible. Determine the expresses for the following quantities in terms of M, X, D, h, and g. Note that these symbols do not include the spring constant. (a) The time elapsed from the instant the block leaves the table to the instant it strikes the floor. (b) The horizontal component of the velocity of the block just before it hits the floor. (c) The work done on the block by the spring. (d) The spring constant. It wasn't given, but I'm sure the springblock is elastic 2. Relevant equations W(elastic) = Fx(cos0) Potential energy = mgh Kinetic energy = .5mv^2 F = kx W = .5kx^2 T = 2pi (sq rt (l/g) 3. The attempt at a solution (a) I think the elapsed time would have something to do with the period. If so, then T = 2pi (sq rt (l/g)) .. and in this case l would be D. So 2pi (sq rt (D/g)) (b) Well, the spring would give M some velocity before it goes off the platform. I think the mgh and 1/2mv2 would have to do something with it, since there has to be kinetic and potential energy. If the potential energy and kinetic energy make up the total energy used, then I can say .5mv^2 = mgh v = sqrt (2gh) (c) The work done is elastic, and should by .5kX^2.. the only problem is I'm not allowed to use k. If k = 4pi^2 x M / T^2, I could just plug in the value I got for T in the T for this equation, then solve for K, and plug that into the work equation.. I think. (d) k = 4 pi^2 x M / T^2 .. so based on my solution for T in the first equation I could then get the answer. 


#2
Dec106, 04:20 AM

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I think you are getting a bit confused about what's going on here. There are two parts to the motion. First, the mass is pushed across the table by the spring. Second, the mass falls to the ground under gravity (and it also has a horizontal velocity). You will need to set up the equations for each part separately.
3(a) I think your equation T = 2 pi sqrt(l/g) is the period of a simple pendulum. I don't see how that is relevant to this problem. When the mass is pushed off the table it falls to the ground under gravity, it is not part of a pendulum. 3(b) I think you are bit coufused here, you seem to be mixing up the KE form the first part of the motion and PE from the second part. Think about each part separately. You might find it easier to work out the problem using K first. Then when you have the answer including K, you might be able to see how you can eliminate it. 


#3
Dec106, 11:12 AM

P: 263

(a) So, perhaps, a = v/t and v = x/t.. so the time would equal
t^2 = D/g = sqrt d/g (b) Could I maybe do the same thing as above.. as in v = x/t.. and substite the h value for x (new length), so the answer would be h / (sqrt d/g) If not, perhaps mgh + .5mv^2 = .5kx^2? At the top there is kinetic energy from the spring since it launched the mass, and it has potential energy on the top. Or maybe I will set it equal to the bottom because Eo = Ef, in which case there is no potential energy,.. so maybe it's .5mvo^2 + mgh = .5mvf^2 ? And I could cancel out an m, and solve for the change in v? And I should change my answers for c and d according to what I get for a and b? 


#4
Dec106, 11:48 AM

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Spring Constant problem
A hint regarding (b): since there is no force acting on the ball during its 'flight' in the horizontal direction, the horizontal component of velocity remains ________.



#5
Dec106, 12:57 PM

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(a) Yes that's the right idea but check the formula. For constant acceleration, v = at, x = ???? (at^2 is close, but wrong)
(b) Read the question carefully, it's asking for the HORIZONTAL component of velocity. See radou's hint above. You found the time in part (a). What is the distance travelled horizontally? (c) and (d) in your original post, you are right, the work done by the spring is (1/2)Kx^2 but you don't know K. When the spring pushes the mass across the table, the work done by the spring is converted into what? Can you write that using the things that you know? 


#6
Dec106, 02:52 PM

P: 263

t = sqrt x/a, which in the terms I'm allowed is sqrt D/g... I dont understand what I'm doing wrong here. (b) Read the question carefully, it's asking for the HORIZONTAL component of velocity. See radou's hint above. You found the time in part (a). What is the distance travelled horizontally? Well, if it remains horizontal without any forces acting on it, then it should be constant. Therefore, v = at, so it would it just be as simple as v = a(sqrt(D/g))? Chances are if I still dont get the t right above, then I wont get this. Another equation I know is Vf^2 = at + Vo^2, which would mean Vf = (sqrt gD) + gt (c) and (d) in your original post, you are right, the work done by the spring is (1/2)Kx^2 but you don't know K. When the spring pushes the mass across the table, the work done by the spring is converted into what? Can you write that using the things that you know? Let me see.. the work is converted into kinetic energy. .5kx^2 = .5mv^2 kx^2 = mv^2 k = mv^2/x^2 I could plug in the value I got for velocity above for v, and then I would get k.. but that's not the work still. Could the work just be .5mv^2, and I'll plug in the value for v as given above.. and then using this I could get the spring constant from the above equation? 


#7
Dec106, 03:09 PM

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No, x = vt only if v is constant, but it isn't constant here.
In the general case dx/dt = v, dv/dt = a, a is a constant acceleration (= g). v = integral from 0 to t of a dt = at x = integral from 0 to t of v dt = integral .... at.dt = ??? Does that make sense? If not, look back to when you did the motion of a particle falling under gravity and see what formula you used. "Therefore, v = at, so it would it just be as simple as ..." yes, that's the way to do it, when you have got the right value of t. (But I think you mean use v = x/t not v = at.). "I could plug in the value I got for velocity above for v, and then I would get k.." Yes that's right. "Could the work just be .5mv^2" ... and yes, again. Actually the work is .5 mv^2 or .5 k x^2, you just said they are both equal, so whichever one you use you will get the same answer. So once you get the formula for (a) sorted out, you are there. 


#8
Dec106, 10:16 PM

P: 263

Okay.. so it all comes down to a then.
If velocity is not constant, then I remember this one formula... Xf = .5at^2 + Vot + Xo If a is constant, then .5at^2 = 0 So X = Vot Vo = X/t t = x/v,hm.. at.dt = v I also know that Vf2 = Vo2 + 2as . s being distance In that case, v = sqrt 2gD Were one of these the equations you were referring to? 


#9
Dec106, 10:42 PM

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Part of this problem is just a projecile problem with an initial height and an initial horizontal velocity. You need to use separate equations for the x motion and the y motion that have the form of the equations you wrote. Write the equation that gives the position, x(t), relative to the edge of the table in terms of the initlal velocity. Write the equation that gives the height above the floor y(t) in terms of the initial height and the acceleration (intitial vertical velocity is zero) These equations both involve time, which can be taken as zero when the mass leaves the table. At some point in time, the block hits the floor (y = 0) and at that moment x = D. To get you started, once you have the correct equations, t can be found in terms of h and g. 


#10
Dec106, 10:52 PM

P: 312

Horizontal velocity IS constant because the vectors are perpendicular. Part b is just D/answer to part a
And the conclusion that the work done by the spring is 1/2mv^2 is called the workenergy theorem. Work=deltaKE. The block starts out with no KE and then gains KE equal to 1/2m(answer to part b)^2. But it all starts out with the time it takes to fall a vertical distance h, so start there 


#11
Dec106, 11:42 PM

P: 263

I meant to say that is a is zero, a is constant, I just got it mixed up a bit x_x;
Okay.. position at the end of the table in terms of intial velocity. x = .5at^2 + Vot (x  .5gt^2)/t = Vo x = t (.5at + Vo) x/t – .5at = Vo Equation that gives the height.. V^2 = 2ax x = h v = at h = v^2 / 2a I could just do h = (at)^2 / 2a, which would give at^2 / 2 = h This is in terms of h and g.. T^2 = 2h/g .. t = sqrt 2h/g 


#12
Dec106, 11:44 PM

P: 263

If my work for a is done correctly, then I can finish the problem. 


#13
Dec106, 11:44 PM

P: 263

Oh yeah, and thanks everyone ^_^;; This is really helpful.



#14
Dec206, 07:08 AM

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twotaileddemon, when you post things that look wrong, sometimes it's a bit hard to tell whether you don't understand something or whether you just made a typo. I think I know a bit about maths and phyisics, but mindreading is WAY harder to do than maths ;)
In your post #11 x = .5at^2 + Vot (x  .5gt^2)/t = Vo x = t (.5at + Vo) x/t – .5at = Vo All that is correct, but you missed one thing. What is the initial vertical velocity? (Remember the horizontal and vertical components of the velocity are independent). I don't quite follow your logic after that, but you did finish up with the right equation: t = sqrt 2h/g. You should be OK with the other parts now you have the right value of t. Good luck. 


#15
Dec206, 10:41 AM

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y = y_o + Voyt  ½gt² because the vertical acceleration is g You are free to choose y = 0 any way you want, but the simplest choice is to use the floor as the reference point (y = 0). You made a similar choice when you decided to make x = 0 and t = 0 the point and the moment in time where and when the block leaves the table. The problem statement tells you the values of Voy and y_o, once the reference point is chosen. Choosing y = 0 to be the floor gives y = h  ½gt² You need this equation to find the time when the block hits the floor (when y = 0). This is where t = sqrt(2h/g) comes from. Your velocity equation h = v^2 / 2a needs to be cleaned up. The a in this equation is of course g, but the v is not the total v, nor is it the velocity you were asked to find. It is the vertical component of v and should be represented as such. Let's call it v_y. Then your equation becomes h = (v_y)²/2g where h and g are the known quatities in the problem This equation could be solved for v_y. However, the problem never asks you to find this component of the velocity. 


#16
Dec206, 12:54 PM

P: 263

Okay, so my answer t = sqrt (2h/g) was correct, but I derived it wrong.
Then let me try to understand what you said in my own terms, because even if I have an answer I need to know the correct method to get it. I can see why the vertical acceleration is g. y = y_o + Voty  1/2gt^2 is just the vertical component purely. The initial velocity for the vertical component at the top does not exist, so we can ignore the middle part. y_o just equals the height, the 1/2gt^2 is the the acceleration and time needed during the whole process. Therefore, at the bottom y = 0 because that is the end of the process. So, we can say 0 = h .5gt^2. h = .5gt^2 h = .5gt^2 2h/g = t^2 sqrt 2h/g = t Okay, that makes a lot more sense. And if I were asked to find a vertical component for the velocity I could also do that, just using x components instead. Thanks for your help ^^; I will look over my notes on x and y components because I would like to practice those more. 


#17
Dec206, 07:56 PM

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It will be worth your time to reinforce the ideas of separating the x and y components for projectile motion. You will encounter similar problems that have constant forces from other sources besides gravity later on, and you will need to rely on your understanding of projectile motion to approach all such problems. 


#18
Dec206, 08:31 PM

P: 263

Okay thanks ^^
What I meant is that if I was going to find the y component of velocity, I could not use the x components. I just typed in an x instead of a y by accident.. sorry. 


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