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Stuck on one little thing

  1. Nov 9, 2004 #1
    A mass, m1 = 9.04 kg, is in equilibrium while connected to a light spring of constant k = 122 N/m that is fastened to a wall
    A second mass, m2 = 7.31 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.220 m The system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. Determine the value of v.

    potential energy= kinetic energy
    0.5kA^2=0.5mv^2
    v= sqrt (kA^2/(m1+m2))
    v=0.601 m/s

    How far apart are the masses when the spring is fully stretched for the first time?

    This would involve finding the time that this occurs in order to get the distance. With the variables I have.. i don't know what to do next.. i have speed.. and masses..and an amplitude..Help?
     
  2. jcsd
  3. Nov 10, 2004 #2
    Simple Harmonic Motion, Newton's 2nd Law, Hooke's Law, Law of Conservation of Energy

    New amplitude of the block- m1-spring system :
    [tex]\frac{1}{2}m_{1}v^2=\frac{1}{2}kA^2[/tex]
    A=0.164 m with v is the speed of m1 and m2 at the equlibrium point.
    Simple harmonic motion of the system :
    x=Asinwt with A=0.164 m
    [tex]-kx=m_{1}a[/tex]
    [tex]a=-\frac{k}{m_{1}}x[/tex]
    [tex]a=-w^2x[/tex]
    [tex]w=\sqrt{\frac{k}{m_{1}}}[/tex]
    [tex]x=Asinwt[/tex]
    Left direction : Positive
    When m1 is fully stretched for the first time, x=0.164 m, then
    [tex]sin\sqrt{\frac{k}{m_{1}}}t=1[/tex]
    t=0.428 s
    m1 takes 0.428 s to be fully strecthed relative from the equilibrium point.
    At this time interval, m2 has gone s=ut=0.257 m from the equilibrium point. so the distance between the two masses is 0.257m -0.164 m =0.093 m
     
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