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Changes in Mechanical Energy for non-conservative Forces Problem

  1. Mar 16, 2010 #1
    A toy cannon uses a spring to project a 5.30 g soft rubber ball. THe spring is originally compressed by 5.00 cm and has a force constant of 8 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon and the barrel exerts a constant friction force of .032 N on the ball.

    (a) With what speed does the projectile leave the barrel of the cannon?
    (b) At what point does the ball have maximum speed?
    (c) What is this maximum speed?

    2. Relevant equations
    d = delta/change in

    dK - dPE (due to gravity) - dPE (due to spring) = Wf

    .5m(Vf)^2 - .5m(Vi)^2 + mgh - mghi + .5k(xf)^2 - .5k(xi)^2 = -fd

    (assuming that the only one i need?)

    3. The attempt at a solution

    I've tried going over this a couple of times and have no idea how to get the answer to (a) which is 1.40 m/s. Since there's no PE due to gravity since its all in the horizontal, i can eliminate dPEg, and since it starts at rest i can eliminate Ki. I think i can also get rid of the final x position of the spring since after the cannon fires the spring returns to its unstretched position? which would leave me with:

    .5m(Vf)^2 - .5k(xi)^2 = -fd

    I'm assuming with it being compressed and having xf being 0, that xi = -.05 m and since it travels .15 m, that d = .1 m, not .15 m. since you're given everything else when you rearrange it to isolate Vf you get:

    Vf = sqrt[((k(xi)^2) - (2fd)) / m]

    but when i do this i get Vf = .505 m/s.

    I've gone over it and looked through the book again and again but i have to be missing something. Any help would be appreciated as I have a midterm on the material tomorrow morning. Also, my first post so if its difficult to read I apologize. Thanks.
     
  2. jcsd
  3. Mar 16, 2010 #2
    When I plug in the numbers to the formula you gave, I get 1.40 m/s
     
  4. Mar 16, 2010 #3
    Vf = sqrt[((k(xi)^2) - (2fd)) / m]

    I don't know how but I keep getting .506 m/s...

    k*(xi)^2 = .02

    2*f*d = .0064

    .02 - .0064 = .0136

    .0136 / .053 kg = .257

    sqrt(.257) = .506 m/s....

    You say you're getting 1.40 m/s but you can't with that formula, i've checked and my calculator is in degrees so that's not the problem... I don't know what's up with it. Help??
     
  5. Mar 16, 2010 #4
    2*.032N*15cm = .0096 J

    and 5.3g = .0053kg
     
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