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How far will this block travel when launched on the floor?

  1. Oct 13, 2006 #1
    I consider myself good in physics, but I was unable to figure this problem out after working on it for almost an hour and a half. I approached it many different ways, but I always hit a dead end or a solution I knew was wrong.
    Our teacher gave us a wooden block, a meter, stick, and a timer. He gave us no constants. We were asked to determine how far the block would travel across the floor if it were launched with an initial velocity of 11.3 ft/s on the floor.

    The main problem was we couldn't test the block with different initial velocities because there was no way to measure it. We were given half an hour, and with time running out we decided the best way to do it was launch it with an unknown initial velocity and record the distance and time. Therefore, if the final velocity, displacement, and time are known, theoretically one could solve for acceleration. The problem is I couldn’t no matter what models I used to find acceleration.
    Thinking back on it, if I had throw the block a certain distance and measured that distance and time I could have obtained the initial velocity, and let it slide afterwards to get a displacement and another time. It seems too complicated though. Any help?
     
    Last edited: Oct 13, 2006
  2. jcsd
  3. Oct 13, 2006 #2

    Doc Al

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    Staff: Mentor

    The key, as you seem to realize, is to somehow measure the acceleration. The acceleration, which depends on the friction, will be the same for any speed. What's a simple kinematic relationship between distance, time, and acceleration?
     
  4. Oct 13, 2006 #3
    x = vt + at^2

    I first tried to use that and calculate the quadratic fuction on my calculator. It wasn't right though because its not quadratic if velocity is changing.
     
  5. Oct 13, 2006 #4

    Doc Al

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    Staff: Mentor

    The proper equation is:
    [tex]x = v_0 t + (1/2) a t^2[/tex]
    But if you view the motion backwards, then you can take v_0 = 0.
     
  6. Oct 13, 2006 #5
    Sorry, my last post was sloppy. That was the equation I used.
    But when I did that, my a=9 ft/s/s which gives the distance as about 8 feet. The actual distance was about 45 feet.
    Normally, I would conclude my experimental data is off, but its seems way too off.
     
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