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The last 2 problems

  1. Oct 14, 2003 #1
    I took more time to go through these problems, and I know they're not quite right yet, so if anyone would be willing to tune up my math I'd greatly appreciate it :)

    5) A child's pogo stick stores energy in a spring (k = 2.40 104 N/m). At position A (xA = -0.130 m), the spring compression is a maximum and the child is momentarily at rest. At position B (xB = 0), the spring is relaxed and the child is moving upward. At position C , the child is again momentarily at rest at the top of the jump. (a) Calculate the total energy of the system if both potential energies are zero at x = 0.(b) Determine xC.(c) Calculate the speed of the child at x = 0.(d) Determine the value of x for which the kinetic energy of the system is a maximum. e) Calculate the child's maximum upward speed.

    (a) I figured the KE at point XA is zero so the total energy would be all PE. I used the formula U = mgh to get the potential energy. U = (24.5)(9.8)(-.130) = 31.2. (b) Again, at point XC, I figured KE to be zero. Since I figured out the total energy in part A, I took that into part b to find the new height. 31.2 = (24.5)(9.8)(h). For part (c) I used the same formula and figured the height to be -.130 - 0 since that's the change in height from point a to point b. Then I put that into the formula KE = (1/2)mv^2. I got v = 1.60 m/s. For (d) I took the formula U = (1/2)kx^2. 31.2 = (1/2)(2.40x10^4)(x^2). Finally, for part (e) I took U=mgh and got U = (24.5)(9.8)(-.130+.130).

    6) A block of mass m = 3.50 kg situated on a rough incline at an angle of = 37.0° is connected to a spring of negligible mass having a spring constant of 100 N/m. The pulley is frictionelss. The block is released from rest when the spring is unstretched. The block moves 15.0 cm down the incline before coming to rest. Find the coefficient of kinetic friction between block and incline.

    I know that F = -kx. I got F to be -15. I know the normal force is mg. I got n to be 34.3. So finally I took the formula F=(mu)(n). -15 = (mu)(34.3). That makes the coefficient of friction to be .437. I know I'm missing something because I didn't account for the 37 degree incline, though.
  2. jcsd
  3. Oct 14, 2003 #2
    5(a) is wrong because you haven't considered the spring potential energy.
    Us = (1/2)kx^2 = .5(2.40104)(.13)^2
    Us = .0203 N*m

    That was the spring potential energy at x = -.13 m, which would have been converted to kinetic energy at x = 0 m. (Note that the child is moving upward at that point, so, while the potential energy is 0, the total is not.)

    Also, you seem to be using 24.5N as the mass. Where did that come from?

    Finally, when you did your multiplication, you dropped the minus sign. Can't do that.

    Sorry, I don't have time to do any more with this, but hopefully that should let you get started fixing it.
  4. Oct 14, 2003 #3
    oops, sorry .. I was copying and pasting the questions from web assign and missed the part where it said "Assume that the combined mass of the child and the pogo stick is 24.5 kg." Also, k is actually 2.40e4 N/m. I thought I caught everything, but I guess I missed that.

    But in any case, I tried the spring constant equation and it gave me potential energy equal to 203 J. However, when I entered that into webassign, it's still not right. I'm so frustrated with these problems!
  5. Oct 14, 2003 #4

    When you did that (U = 203 J) you were neglecting the gravitational potential energy, which at x = -.13m is ... (you already computed that, except it's negative).

    Put 'em together & you should have it.

    'Bye. Gotta go to class.
  6. Oct 14, 2003 #5
    Exxcellent. I got part (a) and part (c) from that. So basically the formula is PE = mgx+(1/2)kx^2. I tried to get part (b) by plugging the known PE into the formula again and solving for x, but that didn't work. I got .0838 m. (d) I got to thinking .. and wouldn't the x value for which KE is at a max be zero? I know it's wrong, but since the spring isn't moving at all and PE is zero, it seems like that's where the most KE would be. And as for (e), I think the formula KE=(1/2)mv^2 would be used, but I'm unsure for what to put in for the value of KE.

    If you have any time later and have any words of wisdom on question 2, that would be greatly appreciated as well.

    Thanks a bunch :smile:
  7. Oct 15, 2003 #6
    Not exactly. If you define x as the height relative to point B, yes, mgx is the gravitational potential energy at every value of x. But in the spring "formula", x represents the amount of compression of the spring. To be precise, we should really say that in this particular problem
    Us = (1/2)k(x)2 but only for -.130 <= x <= 0 since this spring only pushes. It never pulls back in the other direction.

    You really have to stop "plugging into formulas" & think about what's happening. There are only 3 positions to consider in this problem:

    If you make a diagram like this it might help clarify your thinking>>

    | C = maximum height; velocity = 0
    | B = origin; potential energies = 0; velocity maximum
    | A = spring compression at maximum; velocity = 0

    C. Position of maximum height. Here, velocity is zero (that's clear, isn't it?) Therefore kinetic energy is zero. The spring is not compressed. Therefore, ALL of the energy is gravitational potential energy.
    U = mgx = 171.587
    x = 171.587/(24.5*9.8) = .715 m

    B. Your problem DEFINES this point as the origin -- the point where x=0, so therefore gravitational potential energy U=mgx=0 here.
    They also tell you that here the spring is relaxed, so therefore Uspring=(1/2)kx^2=0.
    Therefore, here ALL of the energy is kinetic energy.
    Etot = (1/2)mv2
    (1/2)mv2 = 171.587
    v = &radic;(171.587/(2*24.5))
    v = 3.75 m/s

    A. velocity is zero (given), so all energy is potential.
    Etot=Us + Ug
    Etot=(1/2)kx2 + mgh
    Etot=(1/2)(2.4*104)(-.13)2[/sup[] + 24.5*9.8*(-.13)
    Etot=171.587 N*m
  8. Oct 15, 2003 #7
    For the other problem -- it's too complicated to do this using force equations. The force exerted by the spring keeps changing as the degree of compression changes. Also, the point at which the mass stops is not necessarily an equilibrium point. At that point, the coefficient of STATIC friction (which you don't know) comes into play, & if it's not too high, the mass is probably going to bounce & start moving back up. We don't have enough info to determine that.

    But you do have enough info to solve this using ONLY energy considerations, even if that stopping point is just a momentary stopping point before the spring starts pushing the mass back up. Pick either the starting point, or the lowest point -- your choice -- as the origin & come up with expressions for the spring potential energy and the gravitational potential energy at each point. At both points, you know that velocity, and therefore kinetic energy, = 0. Keep in mind that you must take the angle into account to determine the change in GRAVITATIONAL PE. Also, some of the energy is lost -- consider the WORK done by the frictional force while the mass is moving.

    It's not too hard if you do it this way. Give it a try.

    (By the way, where's the pulley? It doesn't seem to belong in this problem.)
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