Mark24688m 0 #1 October 16, 2007 Ok anybody here who is good at physics i need some help. The problem is basically, a sled weighing 20kg is pushed with a velocity of 4m/s up a 20 degree incline with a 0.20 coefficient of friction. How far up the incline does the sled move? If anybody can help me solve this pleeeease do post. Quote Share this post Link to post Share on other sites
pBASEtobe 0 #2 October 16, 2007 I think you're missing something. Either that or you wrote it incorrectly. Are you given a time or maybe a force? Quote Share this post Link to post Share on other sites
jimstermer 0 #3 October 16, 2007 They dont need it. Use the equation vfinal^2 = vinitial^2+2*a*d Sum the forces Ffriction and Fgravity. Ffriction will be 20kg*.2*sin(20)*9.8m/s Fgravity will be 20kg*cos(20)*9.8m/s a will be negative if you make up hill positive so vinitial will be 4, vfinal will be 0, and a will be (Ffriction+fgravity)/20kg. Solve for d Quote Share this post Link to post Share on other sites
Mark24688m 0 #4 October 16, 2007 umm nope thats literally the whole problem out of the book. The force though would just be gravity (g) 9.80m/s/s. Then some kind of trig used either cosine sine or tangent i dont even know if so confused. Quote Share this post Link to post Share on other sites
pBASEtobe 0 #5 October 16, 2007 Oh, he means it was pushed with an INITIAL velocity of 4 m/s. That makes sense now. What he wrote was that it was pushed with a velocity of 4 m/s up a slope. Big difference. Quote Share this post Link to post Share on other sites
Mark24688m 0 #6 October 16, 2007 Quote They dont need it. Use the equation vfinal^2 = vinitial^2+2*a*d Sum the forces Ffriction and Fgravity. Ffriction will be 20kg*.2*sin(20)*9.8m/s Fgravity will be 20kg*cos(20)*9.8m/s a will be negative if you make up hill positive so vinitial will be 4, vfinal will be 0, and a will be (Ffriction+fgravity)/20kg. Solve for d THANKS!!! OK so i got 5.978meters, is that totally fucked or is it just me? Quote Share this post Link to post Share on other sites
jimstermer 0 #7 October 16, 2007 Sure, didnt run the numbers though just make sure you're using the correct sin and cos since its given in degrees and not radians Quote Share this post Link to post Share on other sites
Mark24688m 0 #8 October 16, 2007 ya i made sure it was in degree mode. After substituting everything in the equation i got 0squared = (4) squared + 2(9.87)(d) but Im' kind of stuck on that. (I know, my math is HORRIBLE) After trying to solve that I somehow ended up with 0squared=35.74(d) and thats where i got stuck, 0squared over 35.74 =d Quote Share this post Link to post Share on other sites
jimstermer 0 #9 October 16, 2007 acceleration is negative so: -16=-9.87*2*d Quote Share this post Link to post Share on other sites
Mark24688m 0 #10 October 16, 2007 Hmmm. But then it would end up being -16=2(9.87)(d) ---> -16 = 19.74(d) -----> d= -0.81meters ....but that just doesn't sound right at all. Quote Share this post Link to post Share on other sites
jimstermer 0 #11 October 16, 2007 the acceleration was negative the answer should be positive. Can someone else verify I'm on the right track or the answer is right? Quote Share this post Link to post Share on other sites
kallend 1,651 #12 October 16, 2007 Are you being graded on this homework?... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
Mark24688m 0 #13 October 16, 2007 Quote the acceleration was negative the answer should be positive. Can someone else verify I'm on the right track or the answer is right? Thanks for the help Jim. After I looked it over again it seems to be right. I really appreciate it! Its not really being graded, just homework basically. This question was just reaaaallly bothering me because it seemed so simple, and I couldn't figure it out. Quote Share this post Link to post Share on other sites
