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, and Power problems

  1. Jan 17, 2008 #1
    Ok, this is the last day I have to work on this stuff, so please bear with me. I'll be posting questions in this thread as needed, so check back in every once in a while please.

    A pole vaulter clears 6.00 m. With what velocity does the vaulter strike the mat in the landing area? (Disregard air resistance. g = 9.81 m/s^2)
     
  2. jcsd
  3. Jan 17, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    At 6.00m what is the vaulter's gravitational p.e.? Since you are disregarding air resistance, what is all the energy going to be converted to?
     
  4. Jan 17, 2008 #3
    But PE = mgh, and I don't have m.
     
  5. Jan 17, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    While that is true, as the vaulter falls, kinetic energy is gained while potential energy is lost. What does that mean now?
     
  6. Jan 17, 2008 #5
    mgh = 1/2 mv^2. Masses cancel. Thanks.
     
  7. Jan 17, 2008 #6
    New problem.

    A 40.0 N crate starting at rest slides down a rough 6.0 m long ramp inclined at 30.0 degrees with the horizontal. The force of friction between the crate and the ramp is 6.0 N. Using the work-kinetic energy theorem, find the velocity of the crate at the bottom of the incline.
     
  8. Jan 17, 2008 #7

    rock.freak667

    User Avatar
    Homework Helper

    Well at the top of the 6m incline the crate has PE, and as it slides down all the PE is converted to KE and work done in overcoming friction.
     
  9. Jan 17, 2008 #8
    work done in overcoming friction?
     
  10. Jan 17, 2008 #9

    rock.freak667

    User Avatar
    Homework Helper

    Yes, if there is not sufficient energy to overcome friction, then the block won't move. But the work done in overcoming friction is just simply Force*Distance
     
  11. Jan 17, 2008 #10
    And force is the downward force and distance is 6.0 m?
     
  12. Jan 17, 2008 #11

    rock.freak667

    User Avatar
    Homework Helper

    The frictional force is the opposite direction of the motion of the crate.So it is the frictional force*distance(6)
     
  13. Jan 17, 2008 #12
    The frictional force you speak of is 6.0 N, right?
     
  14. Jan 17, 2008 #13

    rock.freak667

    User Avatar
    Homework Helper

    Indeed
     
  15. Jan 17, 2008 #14
    Ok, I did this:

    mgh=PE
    PE=(40.0)(9.81)(3)=1177.2 J

    I got the height by this:

    6.0 * sin(30)=3

    then:

    PE=KE+W
    PE-W=KE
    W=F(6.0 N)*d(6.0 m)=36 J
    1177.2-36=1141.2 J

    That might be where I messed up, but I kept going:

    KE=1/2mv^2
    2KE=mv^2
    2KE/m=v^2
    (2(1141.2))/40.0=v^2
    57.06=v^2

    So, the square root of 57.06 equals velocity, but it doesn't match the answers I have. It is a multiple choice question, here are the anwers:

    A) 8.7 m/s

    B) 3.3 m/s

    C) 4.5 m/s

    D) 6.4 m/s

    What did I do wrong?
     
  16. Jan 17, 2008 #15

    rl.bhat

    User Avatar
    Homework Helper

    Here m is not equual to 40. mg = 40N. SO m = 40/9.81 kg
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work, Energy, and Power problems
Loading...