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Spring and tension problem

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data
    An ideal spring of unstretched length .20m is placed horizontally on a frictionless table as shown above. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0kg. The 8.0kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m = 4.0kg hangs from the other end of the spring. When this spring-and-blocks system is in equilibrium, the length of the spring is .25m and the 4.0kg block is .70m above the floor.

    (a) On the figures below, draw free-body diagrams showing and labeling the forces on each block when the system is in equilibrium.

    (b) Calculate the tension in the string.

    (c) Calculate the Force constant of the spring

    **The string is now cut at point P, a given location on the horizontal frictionless surface between block M and the pulley. **

    (d) calculate the time taken by the 4.0 kg block to hit the floor.

    (e) Calculate the frequency of oscillation of the 8.0kg block.

    (f) Calculate the meximum speed attained by the 8.0kg block.


    2. Relevant equations
    net F = 0
    f = kx
    y = 1/2gt^2
    f = cycles/sec


    3. The attempt at a solution

    a) For the 4kg block, i said there was an mg downward and a tension upward.
    For the 8kg block, I said there was an mg downward, a normal force upwards, a tension to the right, and a spring force to the left.

    b) F = 0
    W = T
    40N = T

    c) F spring = T
    kx = 40N
    k(.25) = 40N
    k = 160 N/s

    d) y = 1/2gt^2
    .7m = 1/2(9.8m/s^2)(t^2)
    t = .374s

    e) i know that the formula is f = cycles/sec but i'm a little confused on how exactly i can determine the time for a cycle.

    f) PE spring = KE block
    1/2kx^2 = 1/2mv^2
    1/2(160N/s)(.25m)^2 = 1/2(8kg)v^2
    v = 1.12 m/s


    If you could give me some help with e, as well as check over my answers briefly, i'd greatly appreciate it.
     
  2. jcsd
  3. Jan 25, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Good. (You are using g = 10 m/s^2 here.)

    Careful! In Hooke's law, F = kx, x is the amount the spring is stretched, not the length of the spring.

    Good. Here you are using g = 9.8 m/s^2. (Recheck that third digit.)

    That's just the definition of frequency, not how to find it. What you need to understand is simple harmonic motion with springs, and how to calculate the frequency of a mass oscillating on a spring. See: Simple Harmonic Motion

    You'll need to redo this with the correct value for spring stretch.
     
  4. Jan 25, 2007 #3
    Thanks for you help Doc. I fixed the value for the spring constant and got 800N/s. I also reviewed my notes on springs, and found that period is equal to 2(pi)√m/k That would mean T = 2(pi)√8kg/800N/m = .6283s -> f = 1.59 Hz. Does that sound right?
     
    Last edited: Jan 25, 2007
  5. Jan 25, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That sounds about right. My only comment would be to make up your mind as to which value you are using for g; I would use 9.8 m/s^2, not 10.
     
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