1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work Energy Theorem, Kinetic Energy, and Tension

  1. Oct 16, 2009 #1
    1. A rescue helicopter lifts a 90 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 10 m. What is the tension in the cable and how much work is done by the tension in the cable? Use the work-energy theorem to find the final speed of the person as well.


    2. Work Energy theorem: Wnet = (1/2 mass velocity final^2) - (1/2 mass velocity initial^2)
    Tension: T = mass gravity



    3. Tension: mass gravity
    T = 90 kg * 9.8 m/s^2
    T = 882 N
    Which is wrong according to Webassign.

    WET = Delta K
    WET = (1/2 mass velocity final^2) - (1/2 mass velocity initial^2)
    Not sure how to go about solving this part. I guess the masses cancel, but what next?
     
  2. jcsd
  3. Oct 16, 2009 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's no surprise that that is wrong. Remember that the person is accelerating upwards. This means that there is a net upward force on him. If the tension is only just balancing his weight, then obviously the net force is zero, and he won't be accelerating. If that's not clear, then you need to review Newton's 2nd Law.
     
  4. Oct 16, 2009 #3
    Ok, so I tried using Newtons Second Law, Force = mass * acceleration, and now I got 63 N, which is apparently still wrong.
     
  5. Oct 16, 2009 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, that's the NET force which is pulling him upwards. Draw a free body diagram for the person, and you'll see that two forces are acting on him, namely his weight downwards, and the force due to the tension in the cable, upwards. You add the two forces acting on him together (taking direction into account) to get the net (total) force on the person. The upward force (tension) must **exceed** his weight by 63 N in order for there to be a NET upward force of that amount accelerating him upward. Do you understand?
     
  6. Oct 16, 2009 #5
    I think I understand, so would the net force be -9.1 m/s^2 * 90 kg which would equal -819 N?
     
  7. Oct 16, 2009 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No. Did you draw a free body diagram for the person like I suggested?

    T is pulling him upwards.

    Weight is pulling him downwards. Let's call the magnitude of the weight W. Then the weight is -W.

    To find the net force, add the two forces that are acting on him:

    T + (-W) = T - W = Fnet = +63 N

    For T - W to be positive, T must be greater than W. In other words, the force pulling him upwards is greater than the force pulling him downwards, so that the force pulling him upwards "wins." This is what we mean when we say that there is a NET upward force on him. "Net" = the direction and magnitude of the end result, after all forces have been considered.

    T - W = 63 N.

    You know what W is.
     
  8. Oct 16, 2009 #7
    W is equal to the weight, which is the gravitational force times the mass, so -9.8 m/s^2 * 90 kg... which is -882 N. Correct?
     
  9. Oct 16, 2009 #8

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, that's correct. Now, as I've stated before, the strength of the force pulling him upwards must be greater than that (in magnitude) by 63 N.
     
  10. Oct 16, 2009 #9
    So I add 63 N to -882 N to get -819 N? Or is it positive 819 N since it is in the opposite direction of the weight? Or would it be 945 N since that is what 63 N + 882N (the opposite of the weight) equals?
     
  11. Oct 16, 2009 #10

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There is only one of those possibilities that makes sense (the third one that you wrote). I am going to tell you the reason for the fourth time: the force pulling him upward has to be larger (in magnitude) than the force pulling him downward in order for him to be moving upward.

    One way to think of it:

    T - W = 63

    T - 882 = 63

    T = 882 + 63

    Just think of it in terms of magnitudes (ignoring direction). The *magnitude* of the weight is 882 N, and the magnitude of the tension is 63 N larger than that.
     
  12. Oct 16, 2009 #11
    So the work done by the tension would be equal to the force times the displacement, or 945 N * 10 m, which would equal 9450 J?
     
  13. Oct 16, 2009 #12

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sounds about right to me!
     
  14. Oct 17, 2009 #13
    So then to find the final speed of the person I'm supposed to use the work energy theorem which will help me find the change in kinetic energy which is [ (1/2 mass velocity final^2) - (1/2 mass velocity initial^2) ]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Work Energy Theorem, Kinetic Energy, and Tension
Loading...