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!

Homework help

  1. Oct 4, 2005 #1
    I'm having problems with the following two problems. Thank you for any help!
    72) A skiier coasts down a very smooth 10-meter high slope. If the speed of the skiier on the top of the slope is 5.0m/s, what is his speed at the bottom of the slope?
    I calculated 2gy= v2 and plugged in v= sq root (2x 9.8m/s2 x10m)= 14m/s Does this look correct? We did an example like this in class but we used vinitial = 0 m/s where its 5.0m/s in this problem so I dont know how to go on from there!
    79) In #72, if the skiier has a mass of 60kg and the force of friction retards his motion by doing 2500 J of work, what is his speed at the bottom of the slope?
    I know that there is an equation v= sq root (vo2 - 2ukmgd) but I dont know where to plug #s in or if this is even the right equation?!
    Thanks again!
     
  2. jcsd
  3. Oct 4, 2005 #2

    Päällikkö

    User Avatar
    Homework Helper

    The principle of conservation of energy:
    [tex]K_A + U_A \pm W = K_B + U_B[/tex]

    Can you take it from here?
     
  4. Oct 4, 2005 #3
    No- that makes me more confused!
     
  5. Oct 4, 2005 #4

    Päällikkö

    User Avatar
    Homework Helper

    Oh, sorry :).
    Well the above equation translates in this case into:
    [tex]\frac{1}{2}mv_A^2 + mgh_A \pm W = \frac{1}{2}mv_B^2 + mgh_B[/tex]

    Does this help at all?
     
  6. Oct 4, 2005 #5
    ok- i can figure out # 72 now but How do I apply the "force of friction retards his motion by doing 2500 J of work" in #79>?
     
  7. Oct 4, 2005 #6
    Also, I calculated #72 out, and I'm getting 5.0 m/s but I dont understand how the speed is the same?!
     
  8. Oct 4, 2005 #7

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    OK, let's back up here.

    No, it is wrong.

    Write down an expression for the total energy at the top of the slope, and another expression for the total energy at the bottom. Then apply the law of conservation of energy, just like Päällikkö said. W represents the work done by nonconservative forces. Since #72 doesn't mention friction, you can set W=0 for that one.

    It is not the correct equation. Again, you have to apply the law of conservation of energy, and this time the work W is not zero.

    Brianna, you seem to be looking for a formula that will solve every problem, instead of thinking about the problem. You can't do physics formulaically. You have to do some analysis wherein you apply physical principles, and in this case the principle is the law of conservation of energy.
     
  9. Oct 4, 2005 #8
    Ok- i'm sorry to keep asking about the same questions but I tried to apply the law of conservation of energy to the two problems and I'm not coming out with the right answers:
    72) I understand that PE at the top of the hill will be zero because the object is already moving, and it will also be 0 at the bottom of the hill because y=0. Applying this, that should leave KEa = KEb: .5mv2a= .5mv2b. This gives me .5(5)2 = .5v2 because m will cancel out. After calculating, I'm winding up with 5 m/s but that can't be the correct answer because the speed should increase as the skiier slides down the hill.
    79) Again, PE should be zero at the top and bottom of the hill but now W is involved. SO I plugged in .5(60kg)(5)2 +2500 J = .5(60kg)(v)2. I calculated this out and got 10 m/s but I was told the answer is 12 m/s
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?