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Homework Help: Gravity and Air Resistance

  1. Jul 15, 2007 #1
    I'm trying to figure out this problem...

    A rock of mass 49 kg accidentally breaks loose from the edge of a cliff and falls straight down. The magnitude of the air resistance that opposes its downward motion is 249 N. What is the magnitude of the acceleration of the rock?

    I know that I should multiply 49kgx9.8 to figure out the effect gravity has, right?, but I have no idea how to figure in the 249 N air resistance...
  2. jcsd
  3. Jul 15, 2007 #2


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    Homework Helper

    um, sure. of course, the accleration due to gravity will be the same for any object regardless of the mass...

    P.S. this forum is not where you post homework questions. there is a specific forum for that.
  4. Jul 15, 2007 #3
    Draw a free-body diagram and label the two forces that act on the rock.

    The force that is exerted on an object of mass m due to gravity near the earth's surface is

    F=mg where g = ~9.81m/s^2

    Then you have the force of 249N acting in the opposite direction (air resistance). Due to the vector nature of these forces, how can you find the resultant force?
  5. Jul 15, 2007 #4
    Would I then subtract 249N from F=mg?
  6. Jul 15, 2007 #5
    Ps, sorry I didn't know questions were supposed to be posted in another forum! In the future, I will post elsewhere! :)
  7. Jul 15, 2007 #6
    Yes that is correct. You first have a adopt a sign convention. Lets call upwards positive and downwards negative. (when you said would I then subtract 249N from f=mg you were basically calling up negative and downwards positive, it doesn't matter as long as you keep it the same throughout the problem. I prefer to call up positive)

    [tex]F_{net}=\Sigma F[/tex]

    =[tex]F_{air resistance}+F_{g}[/tex]

    =[tex]249 N - 480.69N[/tex](since gravity acts downwards we assign the force a negative value)

    =-231.69N or [tex]2.3x10^{2}N[/tex] downwards.

    Now use F=ma to calculate the acceleration
    Last edited: Jul 15, 2007
  8. Jul 15, 2007 #7
    Thank you for your help!
  9. Jul 15, 2007 #8
    Anytime. Good luck.
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