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Help finding work of gravity

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    ASSUME: For all parts of this question, friction is negligible.

    A helicopter is used to lift an astronaut of mass m = 69.2 kg a distance s = 11 m vertically out of the ocean by means of a cable. The astronaut rises with constant upward acceleration of magnitude a = 7.44 m/s2 until she reaches the helicopter.

    Find Wg, the work done by gravity on the astronaut.


    2. Relevant equations
    W=F*distance
    F=mg
    F=ma


    3. The attempt at a solution
    I've come to a couple of conclusions, but I think I'm missing a part or two. Here is what I've done.

    Wg=F*distance=mg*s
    Wg=69.2*9.81*11

    That was wrong.

    Then I tried this:

    If we make gravity the positive direction, then:

    F+mg= (-ma)
    F= (-mg-ma) = -m(g+a)
    Wg= -m(g+a) * s = 13130.7

    That was wrong as well, so I'm trying to figure what I'm doing wrong. I have a feeling the answer is somewhere between these two.

    Also, I know someone is going to say to draw this out. I always do, but I guess I just think differently or improperly for some reason. I have it drawn like this:

    Helicopter is on top, and a person is hanging onto a cable that is 11m long. The person has a mass of 69.2kg, and the acceleration is in the positive y-direction of 7.44 m/s/s. That means that gravity acts in the negative direction, so I guess you could say W= (-mg)*h, but the numerical answer besides the negative sign is still the same.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 3, 2012 #2
    You're over complicating things.

    W = SFdx, or for a constant force W = Fx

    Wgravity = mgh, where m is the mass of the astronaut, g is the acceleration of gravity, and h is the distance traveled. Work will be negative, though, since the force of gravity is acting opposite to the displacement of the astronaut.

    The work done by the helicopter would be a bit more complex though...
     
  4. Mar 4, 2012 #3
    So to find Wgravity, you said it's equal to mgh. Isn't that what I said in my original attempt to solve the problem?

    Wg=F*s=mg*s, where s is the distance (in this case, s is the height). I'm not sure if you were pointing out that I did this correctly or not..maybe partially correct?
     
  5. Mar 4, 2012 #4
    I was pointing out that you over complicated things. Sorry if I over complicated my response LOL :)

    Anyway, here's what I'm trying to say.

    The astronaut originally has 0 energy (0 U and 0 K) when in the water right? The helicopter does work on the astronaut, which is countered by the work done by gravity. It doesn't matter how fast, or with what acceleration the astronaut is moved up by, since the acceleration of gravity is constant. Therefore, answering you, yes you are correct, the work done by gravity is just mgh, where h is deltah, or the change in position.
     
  6. Mar 4, 2012 #5
    EDIT: Sorry, the problem was actually the negative. I didn't realize that direction mattered that much! I thought if we had switched it around so that we made gravity in the positive direction, then the amount of work is the same numerical answer, and we could ignore the negative. Big mistake on my part!
     
    Last edited: Mar 4, 2012
  7. Mar 4, 2012 #6

    gneill

    User Avatar

    Staff: Mentor

    Regardless of your choice of coordinate system, the motion of the object is in the opposite direction to the force being considered.

    You didn't change your significant digits, only their format. 7.47e3 would be a change in significant digits.
     
  8. Mar 4, 2012 #7
    Uh okay, but as gneill said, the displacement is opposite to the work being done on it, therefore the work is negative. I don't know what your teachers want (maybe the magnitude, but thats kinda silly), but regardless, gravity does negative work on the astronaut in this problem.
     
  9. Mar 4, 2012 #8
    Oops :roll: lol You're right.

    I think it was just a bad assumption on my part rather than my professor's fault. Nevertheless, I thank you both for your time and help. :)
     
  10. Mar 4, 2012 #9
    Anytime :) Good luck
     
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