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!

Puzzler Answer

  1. Nov 11, 2003 #1
    We were given the answer to this puzzler and its 71.96875 N
     
  2. jcsd
  3. Nov 11, 2003 #2
    Yes, that's the same answer I got using the method I described to you.
     
  4. Nov 11, 2003 #3
    What I did...

    I will show you what I did and then can you please tell me what I'm doing wrong.

    Consider Vertical:
    ¥Äx= -12.96 m
    Vo= 0 m/s
    a= -9.80 m/s
    v= ?
    t =?

    v©÷ = vo©÷ + 2a¥Äx
    v©÷ = 0 + 2a¥Äx
    = 2a¥Äx
    = 2(-9.80)(-12.96)
    = 254.016
    = ¡î254.016
    =15.93 m/s

    v = vo + at
    15.93 = 0 + (-9.80)t
    t= 1.626

    Consider horizontal
    ¥Äx = 7.5 + 5= 12.5 m
    t = 1.626 s
    v= ?
    ¥Äx = vt
    v= ¥Äx/t
    = 7.688 m/s

    What do I do from here and is this really right?
     
  5. Nov 11, 2003 #4
    The equations in the last one are a little wack. sorry...
    © = squared
    ¥Äx= delta x
     
    Last edited: Nov 11, 2003
  6. Nov 11, 2003 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Re: What I did...

    You calculated the time to fall correctly. But you are using the wrong horizontal distance to calculate the horizontal speed of the box as it falls:

    distance = 7.5m (do not add the distance the box was pushed!)
     
  7. Nov 11, 2003 #6
    Ok...
    so I got vx to be 4.613 m/s. Then I use Pythagoras to find the velocity (vy-squared- + vx-squared- = v-squared) and I got it to be 16.59. Then to find acceleration I went:
    a= V/t
    = 10.20 m/s-squared-

    Am I going right...because then I get F= ma
    F= 10.8 (10.20)
    =110.19 N

    I don't think that's right.....


    By the way...Thanks for all of your help!!

    Oh and do you have a messenger system of some sort cuz that would make this much easier.
     
  8. Nov 11, 2003 #7

    Doc Al

    User Avatar

    Staff: Mentor

    At the moment the box leaves the cliff, Vy = 0. The only speed is horizontal, which you just calculated: Vx.

    So now attack the first part of the motion: the box being pushed along the ground. Tell us everything you know about that segment of the motion, from initial push until the box sails off the cliff.
     
  9. Nov 11, 2003 #8
    Delta X = 5.0 m
    Vx= 4.613 m/s
    t =?

    Delta X = Vx(t)
    t= delta x/ vx
    = 5/4.613
    = 1.083 s

    So do I add 1.083 to the previously establish 1.626 and then get 2.709 for total time?
     
  10. Nov 11, 2003 #9
    You don't care about the total time; you want the acceleration. You know the initial velocity v0=0 m/s, the final velocity v=4.61 m/s, and the distance x=5 m. There is a kinematics formula that relates those three quantities to the acceleration a, without involving time.
     
  11. Nov 11, 2003 #10
    Ok...so I have that. But what is SOOOOO EXTREMELY FRUSTRATING...IS WHAT DO I DO AFTER I HAVE ALL THAT!! The acceleration becomes 0.4163 m/s-squared-. F= ma.... that makes is 10.8(0.4163)=4.49604 N that's no where near the answer.
     
  12. Nov 11, 2003 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Relax....

    First step is to calculate the acceleration properly. I don't think you did. Tell us how you did it.

    Once you find the acceleration, then apply F=ma properly. What are all the forces acting on the box?
     
  13. Nov 11, 2003 #12
    K...I found acceleration wrong. I realized that.
    Acceleration = 1.733 m/s -squared-

    So then do I go F=ma
    F= 10.8 (1.733)
    = 18.7164

    So is that Fnet? If it is...then what?
     
  14. Nov 11, 2003 #13

    Doc Al

    User Avatar

    Staff: Mentor

    How did you calculate the acceleration?
     
  15. Nov 11, 2003 #14
    v-squared- = vo-squared- + 2aDelta X
    4.163 -squared- = 0 + 2(5)a
    17.330569 =10a
    a=1.733
     
  16. Nov 11, 2003 #15
    That's not what I got for acceleration. I got 2.13 m/s2.

    If you get the net acceleration right, then you multiply it by the mass to get the net force, as you described.
     
  17. Nov 11, 2003 #16
    How did you get that for acceleration?
     
  18. Nov 11, 2003 #17

    Doc Al

    User Avatar

    Staff: Mentor

    You have a typo: the value for v is not 4.16, it's 4.61.
     
  19. Nov 11, 2003 #18
    Thank you soooooooooooooo much for everything!!! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!!


    -Paige
     
  20. Nov 11, 2003 #19

    Doc Al

    User Avatar

    Staff: Mentor

    Are we done?
     
  21. Nov 11, 2003 #20
    YES!!!! And I really have to thank you for being patient with my annoyance. I was a little bit annoying and I know it. So thanx.

    -Paige
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Puzzler Answer
  1. Physics Puzzler! (Replies: 1)

  2. Math Puzzler (Replies: 10)

  3. Circuit puzzler (Replies: 1)

Loading...