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Homework Help: Questions dealing with the Second Law of Motion

  1. Jun 4, 2006 #1
    I have a couple homework problems that are causing me some trouble...

    1. The weight of 600g of salami is:
    a) 0.061 N
    b) 5.9 N
    c) 61 N
    d) 5.9 kN

    I found the answer to be a) 0.061N because F=mg so m=F/g which is
    0.6kg/9.8m/sec.squared. The answer is actually b) 5.9 N but I can not figure out how the got this answer. Any insights to this?

    Also...

    2. A 60-kg person stands on a scale in an elevator. How many newtons does the scale read when a) it is ascending at the constant speed of 3.0 m/sec.squared? b) it is desending at the constant speed of 3.0 m/sec. squared?

    I am not sure how to find this out. On one hand, I think that I need to find the acceleration at these speeds. Yet I do not think that there is any acceleration when the speed is constant. Am I sorta on the right track here?

    Thanks for your help!
     
  2. jcsd
  3. Jun 4, 2006 #2
    for 1 isnt w = mg which give you .6kg *9.8 = 5.88 founded up to 5.9
    for 2 what i think is this not sure if you have the answers to checkif it is correct f = ma f= 60 * 9.8 + and - 3 m/s just a guess not sure
     
  4. Jun 4, 2006 #3

    Hootenanny

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    Hi there Mistaken and welcome to PF.

    Well, you have definatly come to the right place.

    I think you are getting a lttle confused here. Is 600g a force or a mass? Is weight a mass or a force?

    Again, I think you getting a little confused here, you are not given any speeds; you are given accelerations, note the untis m.s-2. The scale in the lift actually measures the reaction force exerted by the lift on the perons. I think it would be helpful here if you drew a free body diagram. A good thread which previously described the mechanics is located https://www.physicsforums.com/showthread.php?t=120110".

    Think about summing the forces acting; [itex]\sum F = m\vec{a}[/itex].

    ~H
     
    Last edited by a moderator: Apr 22, 2017
  5. Jun 4, 2006 #4
    Thanks for the help with the first one. It make a lot more sense now. I actually wrote the wrong units down for the second problem. The elevator is ascending and desending at 3.0 m/s not 3.0 m/sec. squared. So is there any force occuring here? I don't think so because there's no change in the velocities which means there is no acceleration. Am I right?
     
  6. Jun 4, 2006 #5

    Hootenanny

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    Spot on, so does the person's weight change?

    ~H
     
  7. Jun 4, 2006 #6
    No, I don't think so.Their weight in newtons would still be 588N (60kg * 9.8m/s. squared) since no other force is acting on their bodies except for the force due to gravity.
     
  8. Jun 4, 2006 #7

    Hootenanny

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    Spot on :smile:
     
  9. Jun 4, 2006 #8
    Yes! Thanks for you help :)
     
  10. Jun 4, 2006 #9

    Hootenanny

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    My pleasure.
     
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