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Energy Problem

  1. Feb 17, 2007 #1
    1. The problem statement, all variables and given/known data
    A 14.0 kg block is dragged over a rough, horizontal surface by a 78.0 N force acting at 20.0° above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300.

    (a) What is increase in internal energy of the block-surface system due to friction?

    (b) Find the total change in the block's kinetic energy.


    2. Relevant equations

    (a) Fk=(coefficient of kinetic friction)(mass)(gravity)
    Change in energy=(Fk)(displacement)

    (b) Change in K= Work of other forces - Change in energy

    3. The attempt at a solution

    (a)Fk=(coefficient of kinetic friction)(mass)(gravity)=(.3)(14)(9.8)=41.16
    Change in energy=(Fk)(displacement)=41.16(5)=205.8J

    (b)Change in K= Work of other forces - Change in energy
    = (78.0)(5)(cos20)
    = 366.5 - 205.8=160.7J


    however, these answers are wrong...??? Suggestions??
     
  2. jcsd
  3. Feb 17, 2007 #2
    I can't see anything wrong; does the problem want energy in some non-joule unit?
     
  4. Feb 17, 2007 #3

    hage567

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    In this case, is the normal force equal to only mg? If you draw a free body diagram and sum the forces in each direction, there is an extra component in the y direction due to the applied force.
     
  5. Feb 18, 2007 #4
    No, all of the answers need to be in joules. I'm not sure what is wrong either....
     
  6. Feb 18, 2007 #5

    hage567

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    Did you try my suggestion of drawing a diagram with all the forces? You will see there are three forces in the y direction, and I think you need to consider all of them to find the normal force. The normal force will not be just mg, due to the y component of the applied force. Once you get this new value for the normal force, you can find the new value for the frictional force. Then you need to find the component of the applied force in the x direction.
     
  7. Feb 18, 2007 #6
    The normal force for this problem is W=F*displacement*cos90. W=366.5*5.00*0. Which equals zero. Then, I am not sure what to do to find the increase in internal energy and the total change in the block's kinetic energy. That is where my confusion comes to play.
     
  8. Feb 18, 2007 #7

    hage567

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    So the normal force is is not doing any work, because the block is not moving in the y direction. But the normal force is needed to find the frictional force, right? You said before your answers were wrong. I believe it is because you have not calculated the normal force properly.
     
  9. Feb 18, 2007 #8
    My normal force was correct when I entered my answer into webassign. The normal force of 0 was correct. I did that part correctly.
     
  10. Feb 18, 2007 #9

    hage567

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    If the normal force is zero then I don't see how there can be a frictional force. Because to me no frictional force means no loss of kinetic energy. Oh well, I guess maybe I'm not understanding what the problem is really asking. :confused:
     
  11. Feb 18, 2007 #10
    Just looking at this quickly, I see that N = mg + Fsin(20) = 14g + 78sin(20)...

    Then, as you're aware, [tex]F_{k}=\mu N[/tex] and [tex]W_{friction} = F_{k} \cdot D[/tex]
     
  12. Feb 18, 2007 #11

    hage567

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    This is what I meant when I was suggesting you needed to sum the forces in the y-direction to get the normal force. This the same approach I took to solve this problem.
     
  13. Feb 25, 2007 #12
    hola

    do you guys go to SDSU by any chance? hehe im stuck on the same dang problem :eek:
     
  14. Feb 25, 2007 #13
    stuck

    hey so if you do N = mg + Fsin20... then for the change in internal energy due to friction its just deltaE = Uk*N right? wasnt correct... two more tries crappy...
     
  15. Feb 25, 2007 #14

    hage567

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    Uk*N is a measure of force, not energy. Think of the definition of work.
     
  16. Feb 25, 2007 #15
    still wrong

    ok i made that mistake... this is what i have:

    deltaE = fk*d = Uk*N*d = Uk*mg*Fsin20*d...

    still wrong... last try... pulling hair outt arghh:eek:
     
  17. Feb 25, 2007 #16

    hage567

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    Your expression for N is not correct. Can you show how you got to that? Did you draw a free body diagram of all the forces on the block? If you do that, you should see what N really is.
     
  18. Feb 25, 2007 #17
    well i had N = mg but then someone else said you add in the y component of the force Fsin20...
     
  19. Feb 26, 2007 #18

    hage567

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    Yes you have to add it in. That's not what your equation was showing. Did you make some typos?
     
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