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Position of a spring

  1. Oct 9, 2004 #1
    position of a spring!!

    Okay, Here's the problem
    The block in Figure 7-11a (Figure not important) lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.8 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached. (a) What is the position of the block? (b) What is the work that has been done on the block by the applied force? (c) What is the work that has been done on the block by the spring force?

    (a) I drew a free body diagram and it seems like when the block stops the spring force catches up to the applied force counteracting it and stopping the motion of the block. So, Hooke's law states that F(spring) = -kx; since F(spring) = F(applied) --> F(applied) = -kx.
    I crunched those numbers and got -.08 which made sense because the applied force is going to be positive [f(s) is negative in Hooke's law and x being a negative number will cancel the negatives].
    BUT THAT'S WRONG what's the problem!!! :surprised
     
    Last edited: Oct 9, 2004
  2. jcsd
  3. Oct 10, 2004 #2
    Does the block stop WHILE the force is still acting? If this is so then there is a point at which the net force on the block is zero. That means the friction (which will be opposite to instantaneous velocity), the spring force and the applied force will give rise to a net zero force at what you call the "stopping point". Rest of the problem seems pretty standard to me.
     
  4. Oct 10, 2004 #3
    The force is negative, so it's -2.8N and you should get 0.08 m. Then you can find the work which is negative too.
     
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