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Help with Work (kinetic energy?)

  1. Oct 13, 2006 #1
    Starting from rest, a 4.5 kg block slides 2.5 m down a rough 30.0° incline. The coefficent of kinetic friction between the block and the incline is µk = 0.436.

    (a) Determine the work done by the force of gravity. J

    (b) Determine the work done by the friction force between block and incline. J

    (c) Determine the work done by the normal force.J

    I am not sure how to go about solving this and the teacher decided to not explain half the things saying that we dont really need it but she goes and post question on things she decided to skip.

    I know that i would use this formula W = (FcosTheta)DeltaX but not so sure.

    Any help at all would be very helpful.
     
  2. jcsd
  3. Oct 13, 2006 #2

    quasar987

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    You have to find the direction of motion (in the case the parallel to the incline) and the component of the force vector along this direction. Once you have that, the work done on the block as it slides down the incline is defined as the product of this force component with the total distance travelled by the block.

    So start by identifying the forces acting on the block. Then find their components in the direction of motion. Then add them. Then multiply by 2.5m

    Makes sense?
     
  4. Oct 13, 2006 #3
    I have no clue what you just said.....i assume there is F->, <-f, and normal force acting on it but not sure how i go about getting that.
     
  5. Oct 13, 2006 #4

    quasar987

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    You know that forces are vectors right? It is a fact that any vector can be decomposed as a sum of two vectors. Say you have a vector [itex]\vec{F}[/itex]. "Finding the component of [itex]\vec{F}[/itex] along a direction" means that you first have to find a vector parallel to the given direction (let's note it [itex]\vec{F}_{para}[/itex]), and one perpendicular to it (let's note it [itex]\vec{F}_{perp}[/itex]), that sum to [itex]\vec{F}[/itex]. Then, the component of [itex]\vec{F}[/itex] along the given direction is the norm of [itex]\vec{F}_{para}[/itex].

    Let me guess, I just made it worse?
     
    Last edited: Oct 13, 2006
  6. Oct 13, 2006 #5
    I know that forces are vectors and i understand F along a directiong other than that i dont understand what you mean by parallel and perpendicular.
     
  7. Oct 13, 2006 #6

    quasar987

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    By parallel I mean "parallel to the direction". The simplest exemple is when you have a vector epressed in cartesian form:

    [tex]\vec{F}=F_x \vec{i}+F_y \vec{j}[/tex]

    Then [itex]F_x \vec{i}[/itex] is the vector parallel to the x-axis, and just [itex]F_x[/itex] is the component of [itex]\vec{F}[/itex] along the x axis.
     
  8. Oct 13, 2006 #7
    okay i understand it a bit more now. I am looking at the problem but from the given information I am not sure how to find F. Even finding F i dont understand how i would go about getting a, b, c. I email the teacher for clarity also and she did not give any helpful information at all. she said dont look for a formula but make one.
     
  9. Oct 13, 2006 #8
    also to get F is it the same as using F = ma where a = (uk)(g) which is
    a = 4.27

    i got that because uk = -(a/g)
     
    Last edited: Oct 13, 2006
  10. Oct 13, 2006 #9

    quasar987

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    Start by verbally enumerating the forces acting on the block.

    There is...

    1° ?
    2° ?
    3° ?
     
  11. Oct 13, 2006 #10
    T.T <---crying.....Physics is hard to understand. I dont understand what you mean by those degree
     
  12. Oct 13, 2006 #11

    quasar987

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    It's just a hint that there are 3 types of forces acting on the block. I might has well have written

    1)
    2)
    3)

    in an attempt to encourage you to list those forces.
     
  13. Oct 13, 2006 #12
    normal force
    gravity (i think)
    friction force f_k
     
    Last edited: Oct 13, 2006
  14. Oct 13, 2006 #13

    quasar987

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    Good. What is the magnitude of each force? For instance, F=mg

    N=? (normal)
    f=? (friction)

    Also, at this point you should have draw a big picture of the block on the incline, and all three force vectors "emanating" from the block's center.
     
  15. Oct 13, 2006 #14
    N = 44.1cos30?
    f = u_k(n)?
     
  16. Oct 13, 2006 #15

    quasar987

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    If by n you mean N, then yes, that is correct.

    Now take an x-y coordinate system in which the x-axis is parallel to the incline (just take a regular coordinate system and rotate it by 30 degrees)

    Do you agree that finding the components of the forces in the direction of the motion means finding the components in the direction of the negative x-axis?
     
  17. Oct 13, 2006 #16
    Is that a trick question...but I think no.
     
  18. Oct 13, 2006 #17

    quasar987

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    Sorry for leaving you in the middle of the problem like this but I gotta go to sleep.

    You're supposed to know how to find the components of a vector along an axis once you know the angle btw the vector and that axis. (You did it implicitely when you said N=mgcos30°... that was finding the component of gravity along the negative y-axis (as we've defined it) and inversing its sign)

    Once you have the components in the direction of the negative x-axis, just add them and multiply the result by 2.5. That is the work.
     
    Last edited: Oct 13, 2006
  19. Oct 13, 2006 #18

    quasar987

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    It was not a trick question, but I don't know how you drew the incline and if you interpreted my directions correctly for defining the cartesian coordinates.

    Hopefully for you, one of the mentor like Doc Al will help you with the rest. They are more experienced and have more ease then me to explain these concepts in simple undrstandable terms. Sorry I couldn't be of more help.

    And good night. Zzz.
     
  20. Oct 13, 2006 #19
    okay, thanks alot for all the help. I will work on it tomarrow night and see what i get. I will post all work here if i still get the wrong answer.
     
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