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!

Spring & Projectiles

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A spring haveing a force constant k is compressed a distance a frots its natural length, then used to launch an ice cub of mass m up a ramp of length l oriented at an angle theta to the horiaontal. The cube starts form rest at ground level, slides without friction up the ramp and sals off the end. Find the horizontal distance d from the ramp to the point of contact. Neglect Air resistance

    2. Relevant equations



    3. The attempt at a solution

    .5ks2 = .5mv2 - mgh

    v = ((ks2 + 2mgy)/m)

    do i just plug this into x = vxt, or do i have to use some y compnents as well.
     

    Attached Files:

    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    I am pretty sure I see what you're getting at but I think there might be a few typos in your post so I may be wrong.

    You are on the right track so far, but double check your equation for V. I think you might have left out a squared/square root. Also, you seemed to have swapped the h with a y here which is a little confusing as h was previously implied to be the height of the ramp and y would seem to be the height of the object.

    You will have to use the theta of the ramp to break the above V into its x and y components. You can use x = vxt, but you will need to use Vy, h and g to determine t. You might find the following equation helpful:

    y=-.5gt2+ Vyt + h

    y=height of the object
    h=height of the ramp
    g=+9.81m/s2 (your previous equation implied a positive g so I made g positive here)
     
    Last edited: Nov 8, 2009
  4. Nov 9, 2009 #3
    yes v = ((ks2 + 2mgy)/m).5


    t = -vy + [tex]\sqrt{v_y ^2 - 4(-.5g)(y)}[/tex]/-g

    x = vxt = [tex]\sqrt{(5s^2 + 2mgh)/m}[/tex]cosQ[-[tex]\sqrt{(5s^2 + 2mgh)/m}[/tex]sinQ +- [tex]\sqrt{((ks^2 + 2mgh)/m)sin^2 Q + 2gy}[/tex]/-g
     
  5. Nov 9, 2009 #4
    Yep, it is a little hard to tell with the formating but it looks like you got it for the most part. It might just be a formating thing but the only thing I'm not following is the minus sign before the last square root symbol.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spring & Projectiles
  1. Spring and Projectile (Replies: 5)

  2. Spring and projectile (Replies: 2)

Loading...