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The ball's maximum speed relative to the vehicle?

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 226 N/m . The vehicle has a steady acceleration of 5.00 m/s2, and the ball is not oscillating. Suddenly, when the vehicle's speed has reached 45.0 m/s, its engines turn off, thus eliminating its acceleration but not its velocity.

    What will be the ball's maximum speed relative to the vehicle?

    2. Relevant equations
    v = ±ω√(A2 - x2)
    ω = √(k/m)


    3. The attempt at a solution

    I solved for ω and got 8.0356 rad/s
    I tried to use the v equation above and set x = 0 and got 0.598m/s, the site told me my answer was not quite right possibly due to a round-off error.
     
  2. jcsd
  3. Nov 20, 2016 #2

    TSny

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    OK, that looks good.
    What value did you use for A and how did you get it?
     
  4. Nov 20, 2016 #3
    I used 0.0774 m for A. I go that by kx=m(a), solved for x and had that equal to A.
     
  5. Nov 20, 2016 #4

    TSny

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    OK, good. When I use your values of A and ω, I don't get your answer of 0.598 m/s. Maybe you plugged in 0.0744 instead of 0.0774?
     
  6. Nov 20, 2016 #5
    I did plug in 0.0744! Thank you! I'm being very clumsy today and making really clumsy mistakes. I looked at what i wrote just now and saw that from 0.0774 i changed it to 0.0744. Many thanks! This kinda confused me though. How do we know when to set the x to 0?
     
  7. Nov 20, 2016 #6
    Did you get 0.622m/s?
     
  8. Nov 20, 2016 #7

    TSny

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    In this problem, you want maximum speed. So, you pick the value of x that will give max speed.
     
  9. Nov 20, 2016 #8

    TSny

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    Yes.
     
  10. Nov 20, 2016 #9
    when x is equal to 0, does that always indicate maximum speed at that displacement?
     
  11. Nov 20, 2016 #10

    TSny

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    For a mass moving only under the influence of a spring, you will get maximum KE (therefore, max speed) when the PE is minimum. Minimum PE for a spring occurs at what value of x?
     
  12. Nov 20, 2016 #11
    at its equilibrium point? x=0?
     
  13. Nov 20, 2016 #12

    TSny

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    Right. At the equilibrium point, the spring is unstretched and there is no potential energy stored in the spring. So, all of the energy of the system must be in the form of kinetic energy.
     
  14. Nov 20, 2016 #13
    That makes more sense now. Thank you!
     
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