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Minimum string tension needed to move the block

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1.50 kg steel block is at rest on a steel table. A horizontal string pulls on the block. The coefficient of static friction of dry steel on steel is Uk=.800, the coefficient of kinetic friction of dry steel on steel is Uk=.600, and the coefficient of kinetic friction of lubricated steel on steel is UkLUBE=.050

    What is the minimum string tension needed to move the block?
    If the string tension is 19.0 N, what is the block's speed after moving 1.30 m?
    If the string tension is 19.0 N and the table is coated with oil, what is the block's speed after moving 1.30 m?


    2. Relevant equations

    F = (Uk)mg

    3. The attempt at a solution

    What is the minimum string tension needed to move the block?
    F = (.8)(1.5)(9.8) = 11.76
    F = 11.8

    If the string tension is 19.0 N, what is the block's speed after moving 1.30 m?
    Since we know we need at least 11.8 to get it to move, we have an additional 7.2N
    So i figured...

    F = ma
    7.2 N = (1.5kg)a
    a = (7.2)/(1.5)
    a = 4.8 m/(s^2)

    Anyways, im a little confused on what to do after i have the acceleration.



    If the string tension is 19.0 N and the table is coated with oil, what is the block's speed after moving 1.30 m?

    I figure i could approach this the same was as the second question but i would basically just need to recompute the minimal tension required to move it.
     
  2. jcsd
  3. Mar 18, 2009 #2

    PhanthomJay

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    Your first answer on the minimum force necessary to move the block is correct, but the other answers are incorrect, because the block is moving. The net force is not 7.2N. Once you get the acceleration, you can get the speed after 1.3 m by using the kinematic equations, but it is easier to use the Work-Energy theorem, if you are familiar with it.
     
  4. Mar 18, 2009 #3
    How would icorrectly calculate the net force since its not 7.2N? Or what should I calculate first? I need the acceleration right?
     
  5. Mar 18, 2009 #4
    Any more hints?
     
  6. Mar 19, 2009 #5
    I dont get why the net force isn't 7.2N if i add the right pushing forcing thats 19N and the left pushing for thats 11.8N (19N - 11.8N = 7.2N ????)
     
  7. Mar 19, 2009 #6
    Now, basically the kinetic friction coefficient can be used to account friction only when the block starts moving. The moment the block starts to move, the static friction reduces to kinetic friction and the opposition to the applied force gets reduced so that the body can accelerate. Keeping this point in mind, you can calculate the block's speed with the help of kinematic equations.
     
  8. Mar 19, 2009 #7
    So instead could i assume with 19N being applied the total net force after traveling 1.3m would be:

    F = (.6)(1.5)(9.8) = 8.82N

    19N - 8.82N = 10.18N

    10.18 N is the total net force?

    Taking this netforce
    A = -(Uk)(g) = -(.6)(9.8) = 5.88

    X1 = -(V^2)/2(-5.88)

    Since we know X1

    1.3 = [-(V^2)]/[2(-5.88)]

    negatives will end up canceling out...

    1.3 = [(V^2)]/[2(5.88)]

    simply multiplication...
    1.3 = [(V^2)]/[11.76]

    more multiplication...

    15.288 = (V^2)

    Square root of 15.288

    3.90 m/s I know the answer is wrong but thats what i've come up with right now...
     
  9. Mar 19, 2009 #8
    What I meant was, you've applied the minimum force to move the block. Let that be the same. instead for the static friction coefficient, now the block's movement would be resisted only by kinetic friction. Evaluate the same equations using the kinetic friction coefficient. Find out the net acceleration and use kinematic relations.
     
  10. Mar 19, 2009 #9
    Isn't that what i did?????????
     
  11. Mar 19, 2009 #10
    Well, that is the idea. See if your calculations are right.
     
  12. Mar 19, 2009 #11
    Everything seems to check out mathematically. Am i using the wrong equation anywhere???
     
  13. Mar 19, 2009 #12
    Yes, that is correct.
    What was that step for?

    Net force = Mass X Acceleration.
    Find out acceleration and solve the rest.
     
  14. Mar 19, 2009 #13
    Alright so if thats the case...

    Acceleration would equal 3.92 (m/(s^2) correct?
     
  15. Mar 19, 2009 #14
    No, what is the mass of the block? 1.5 Kg.
    What is the net force? 10.18 N.
    Calculate acceleration!
     
  16. Mar 19, 2009 #15
    I used 5.18N as the netforce for my calculations earlier, now that i used 10.18N as my netforce my acceleration should be correct since it is: 6.78 m/(s^2) ???
     
  17. Mar 19, 2009 #16
    Tell me if you've got it right. I may be wrong, you know.. :uhh:
     
  18. Mar 19, 2009 #17
    Yes, go on. Solve the rest and tell me if you get it right.
     
  19. Mar 19, 2009 #18
    Got it!
    I used:
    Vf = final velocity, which we are trying to find
    Vi = initial velocity, which is 0?
    a = 6.67m/s
    X = 1.3m

    (vf^2) = Vi + 2aX

    Anyways after a bit of math the final velocity it reaches when it is at 1.3m is: 4.16 m/s
    Which by the way is correct. However, i was supposed to give the answer as 4.20 instead (something about sig figs) but anyways, i got that part. I'm going to do the next part, but that should be easy to do now *i hope*.
     
  20. Mar 19, 2009 #19
    And for the second part basically following the same methods from before, the correct answer is: 5.16
     
  21. Mar 19, 2009 #20
    Ok. Great. You get the ideas right, then everything is okay. You should understand why the kinetic coefficient is used immediately after the block moves. Problems do vary. You may not be able to apply the same thing to each problem.
     
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