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Work and Energy

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Masses A and B, each having a mass of 32.2 slugs, are constrained to move in frictionless slots. They are connected by a rigid, massless rod of length L = 2 ft. Mass B is connected to two massless linear springs, each having a spring constant k = 60 lbF / ft. The springs are unstretched when the connecting rod to masses A and B is vertical.

    The accompanying schematic:
    https://drive.google.com/file/d/0B2-REaK_0ruSZ3cwV2dZS1VuNVE/view?usp=sharing

    If B is given a slight nudge, what are the velocities of B and A when A descends a distance of D = 3 inches?

    2. Relevant equations
    W1->2 = ΔKE + ΔPEG + ΔPEE

    3. The attempt at a solution
    Masses: mA = mB = m
    Take upward to be positive y and right as positive x.

    At position 1:
    Velocity of A: vA1 = 0
    Velocity of B: vB1 = 0
    Height of A: hA1 = L
    Height of B: hB1 = 0
    Displacement of left spring: xL1 = 0
    Displacement of right spring: xR1 = 0

    Finding the displacement of each spring:
    (L - D)2 + x2= L2
    x = 11.62 in. = 0.9682 ft -> symmetric for both springs

    At position 2:
    Velocity of A: vA2 = ?
    Velocity of B: vB2 = ?
    Height of A: hA2 = L - D
    Height of B: hB2 = 0
    Displacement of left spring: xL2 = x
    Displacement of right spring: xR2 = x

    Work:
    W1->2 = 0 because there are no non-conservative forces acting on the system

    Change in kinetic energy:
    ΔKE = (1/2) * mA * (vA22 - vA12) + (1/2) * mB * (vB22 - vB12)
    ΔKE = (1/2) * m * (vA22 + vB22)

    Change in gravitational potential energy: g = 32.174 ft/s2
    ΔPEG = mA * g * (hA2 - hA1) + mB * g * (hB2 - hB1)
    ΔPEG = m * g * ( - D) = -259.001 slug⋅ft2/s2 = -259.001 ft⋅lbF

    Change in elastic potential energy:
    ΔPEE = (1/2) * k * (xL22 - xL12) + (1/2) * k * (xR22 - xR12)
    ΔPEE = (1/2) * k * ( 2 * x2)
    ΔPEE = k * x2 = 56.245 ft⋅lbF

    Relating vA2 and vB2:
    Take the origin at B's initial position.
    x ≡ horizontal distance from origin to B
    y ≡ vertical distance from origin to A

    x2 + y2 = L2 = constant
    2 * x * (dx/dt) + 2 * y * (dy/dt) = 0
    2 * x * vB + 2 * y * vA = 0
    x * vB + y * vA = 0

    At the instant when A has moved down 3 inches:
    vB2 = [ - (L - D) / x] * vA2 = -1.807 * vA2

    Combine all equations:
    W1->2 = ΔKE + ΔPEG + ΔPEE
    0 = (1/2) * m * [vA22 + ( -1.807 * vA2)2] - 259.001 + 56.245
    vA2 = 1.718 ft/s down
    vB2 = 3.105 ft/s right

    This was a midterm question. Physics enthusiasts, please give feedback my thought process. Much appreciated.
     
  2. jcsd
  3. Oct 29, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!

    Nicely organized. Your work looks correct to me.
     
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