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!

Newton's law question

  1. May 25, 2005 #1

    404

    User Avatar

    A sled of 6.0 kg mass is moving along a smooth, horizontal ice surface with a velocity of X. A force of 36N is applied to the sled in its direction of motion, increasing its velocity to 2X while it moves 10m. Find the sled's original velocity, x. And the length of time that the force acted.

    Smooth surface means frictionless, and the answers should be 6.3m/s and 1.1 sec respectively. Thanks.
     
  2. jcsd
  3. May 25, 2005 #2
    Here are the eqs that you would use
    f =ma
    Vf= Vi +at
    d= Vi(t) + .5a(t^2)
     
  4. May 25, 2005 #3

    404

    User Avatar

    is there a way to solve it w/o d= Vi(t) + .5a(t^2)?

    Our teacher forbid us to use it. We can only use the basic ones like a=delta v/ delta t, v average = delta d / delta t.
     
  5. May 25, 2005 #4
    Why dont you tell us what you think so we can tell you what we think of what you think.
     
  6. May 25, 2005 #5
    what whozum said, you have to show us your work for us to figure out what you did wrong or right. It would be unfair if we just gave you the answer, but those eqs work as well.
     
  7. May 25, 2005 #6

    404

    User Avatar

    Well there's like 20 questions in total, this is the only one out of the bunch I didn't get. I tried to average velocity formula

    like...

    I found acceleration is 6 m/s^2, by using 36N/6kg

    v average = (x + 2x)/2 So velocity average is 1.5x, and I put it back into the velocity average = d / t formula, since i know d, I found t to be 6.67 s, then plugged it into the acceleration formula, 6 = delta v / 6.67 (time) and basically the embarrassing result is nothing close to the answer...

    That was only the first part, never tried to find time as this answer is not correct.
     
  8. May 25, 2005 #7
    Do you knwo the equation for final velocity without knowing time? It involves just acceleration distance and velocity.
     
  9. May 25, 2005 #8
    Your acceleration is correct.
    t=6.67 s is not correct
    So you put 1.5x = 10/t and got t= 6.67s where from?
    You have two variables therefore t= 10/(1.5x)
    now plug that t into your acceleration formula and you will get your speed.
     
  10. May 25, 2005 #9

    404

    User Avatar

    Well we are forbiden to use those equations, only the simple ones.

    And I did plug it in, the variables simpled cancelled out for me. I tried 6 = (2x-x)/t
    t=(2x-x)/6 so i set that equal to (2x-x)/6 = 10/1.5x... (I'm still working at the question though...)
     
  11. May 25, 2005 #10
    (2x-x)/6 = 10/1.5x is the right eq. just a matter of algebra now.
     
  12. May 25, 2005 #11

    404

    User Avatar

    If I'm not mistaken, that comes to

    x/6 = 10/1.5x

    The x's would cancel...
     
  13. May 26, 2005 #12
    Nope, multiply both sides by 1.5x
     
  14. May 26, 2005 #13

    404

    User Avatar

    Lol...Oh jeeze. My bad, I don't know what gotten over me, SO obvious, yet I somehow missed it.

    EDIT: Got both answers now, thanks guys.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newton's law question
Loading...