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What distance does the block slide?

  1. May 24, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-5-24_16-24-33.png

    2. Relevant equations
    m*v^2 / 2 and k*x^2 / 2 and 0,5*m*v^2

    3. The attempt at a solution

    in task a I got correct answer there v = 20 m/s
    in task b I got wrong answer I got the distance = 0,76 m but I should be 32 m

    here is my answer :
    0,5*m*(sin(30)*v) - mu * m*g*cos(30) = m*g*h
    h=0,38 m
    sin (30) = h/L there L is the distance of the block when it goes up
    L= h/sin(30) = 0,76

    What I do wrong in task b ?

    Thanks

     
    Last edited by a moderator: May 24, 2016
  2. jcsd
  3. May 24, 2016 #2
    sin(30)v should be squared
     
  4. May 24, 2016 #3
    I seem to get the answer on the second one to be 35.4012435624
     
  5. May 24, 2016 #4
    Shouldnt this be mu*(m*g*cos(30))*h/cos(30)=mu*m*g*h?
    I think so...
     
  6. May 24, 2016 #5
    What equation you did use ? thanks
     
  7. May 24, 2016 #6
    I tried with 0,5*m*(sin(30)*v)^2 - mu * m*g*cos(30) = m*g*h
    but it does not give me the correct answer
    thanks
     
  8. May 24, 2016 #7
    Now I tried 0,5*m*(sin(30)*v)^2 - mu * m*g*cos(30) *h/sin(30) = m*g*h
    but not working
     
  9. May 24, 2016 #8
    I used a kinematics equation
     
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