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Force in a non-isolated system

  1. Mar 9, 2008 #1
    [SOLVED] force in a non-isolated system

    I don't normally do something like this, but I'm stuck on my homework.

    A lizard jumps, pushing with its legs against the ground. Assume the lizard's inertia is .05 Kg. Its center of mass starts out .5 cm above the floor, and raises 1 cm to 1.5 cm while it is pushing against the floor. At that height, its feet lose contact with the floor and it continues to travel up to a height of .15 m. What is the magnitude of the average force exerted by the floor on the lizard?

    I know that F=ma, but I have no idea how to find the acceleration without the time the action took. I also know that if I have to use the change in center of mass, I'll have to convert the cm to m. But other than that, I'm totally lost.

    Any help would be appreciated.
     
  2. jcsd
  3. Mar 9, 2008 #2

    Doc Al

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    Start by finding the final speed of the center of mass just as the lizard loses contact with the floor. Hint: Use kinematics to find that speed, given the height it reaches.

    Then you can figure out what the force must have been to produce such a change in kinetic energy. (Don't neglect the lizard's weight.)
     
    Last edited: Mar 9, 2008
  4. Mar 9, 2008 #3
    I know that I'm going to sound really stupid, but I still don't get it. I skipped this problem and moved on to others, and did what I could with those, but now I'm just more confused than ever when I look back at this one. :S
     
  5. Mar 9, 2008 #4

    Doc Al

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    Just try it step by step. If you threw a ball straight up and it traveled a maximum distance d above your hand (before falling back down again), could you figure out how fast it must have been moving when it left your hand? Same idea here.

    What's the change in height of the center of mass, from the point where the lizard loses contact with the floor to the point where it reaches maximum height?
     
  6. Mar 9, 2008 #5
    I'm still lost. Every formula I know for finding velocity over some distance depends on the time it took to travel that distance.

    The only solution I can come up with looks like this:

    Vfinal=(2gh)^(1/2), where g=9.8 and h=.15-.015=.135

    So, Vfinal=1.63 m/s

    KE=1/2mv^2, where v=1.63 and m=.05 kg

    So, KE=.07 J

    This represents total work done and W=Force(change in distance), so .07=force(.135)

    Force=.52 N

    Is this right at all?
     
  7. Mar 9, 2008 #6

    Doc Al

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    You don't sound as lost as you think!

    Sounds good to me! Although I would not round off at this point.

    If we were a bit more clever, we could have found the KE directly and skipped finding the speed:
    KE = mgh = 0.06615 J

    Since we're interested in the force of the floor on the lizard, we only care about what happens during the time the lizard is pushing on the floor. What distance does the center of mass move during that time? Use that to find the net force on the lizard, then analyze the forces to figure out what the force of the floor must have been.

    You're doing good.
     
  8. Mar 9, 2008 #7
    Ok, so.....

    KE during contact with floor=.05*9.8*.01=.06615 J

    .06615=Force(.01); Force=.49 N during contact with the floor

    This gives us a net force of 1.01 N on the lizard.

    Since force due to gravity while the lizard is in contact with the floor=mg=.05(9.8)=.49 N.

    And assuming that the downward force of gravity is negative, we get a total upward force of the floor on the lizard of 1.01-(-.49)=1.5 N.

    Right, or did I just totally mess it up?
     
  9. Mar 9, 2008 #8

    Doc Al

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    Not sure where you got 1.01 N. You have the equation correct:
    .06615=Force(.01)

    Redo your solution for the net force. (.49 N is the weight of the lizard; you'll use that later.)
     
  10. Mar 9, 2008 #9
    Sorry, must have hit the wrong button on the calculator.

    Force while in contact with floor is 6.615 N, and therefore Net force is 7.135 N.
     
  11. Mar 9, 2008 #10

    Doc Al

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    The net force on the lizard is 6.615 N (upward), and therefore the force of the floor must be 6.615 N + mg.
     
  12. Mar 9, 2008 #11
    So, that gives us a force on the floor of 7.105 N.

    The next part of the question asks about the acceleration of the lizard during the launch time so 7.105/mass=142.1 m/s^2. But doesn't this sound a little too fast?
     
    Last edited: Mar 9, 2008
  13. Mar 9, 2008 #12

    Doc Al

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    OK.

    The acceleration = (net force)/mass.
    The poor lizard needs to toss his body pretty high in the air (compared to his size), so he needs to produce a considerable acceleration to get himself up to speed in that short a distance.
     
  14. Mar 9, 2008 #13
    Ok, I think that I finally get it. No garuntee I'll be able to do a similar problem in the future. But thanks for all your help. I really appreciate it. :D
     
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