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Another question

  1. Nov 16, 2006 #1
    A 2kg block at the top of an inclined plane at a height h=6m above the ground moves down the plane.

    1) Find the value of the velocity when the block is at the bottom of the plane assuming no friction.

    2) With friction present, the final velocity is only 7.0 m/x at the bottom of the inclined plane. Find work done against friction.

    3) Find an algebraic expression for the coefficient of friction in therms of mass (m), the angle (theta), and the work (W), for the block sliding to the bottom of the inclined plane
     
  2. jcsd
  3. Nov 16, 2006 #2
    you can't just post your homework... show you've tried to solve it.
     
  4. Nov 16, 2006 #3
    I have been trying at this stupid problem all day. It was on a test I took, but my whole class couldn't figure out any of the extended response so he let us take it home. For this problem, I kept messing aroudn with the potential energy (mgh) and kinetic energy ((1/2)mv^2) formula but I keep getting really weird answers. I also used the W=energy final - energy initial formulas but that seemed like much of a stretch, and I'm so close to giving up that my eyes are burning but I think that is the point of physics. Any hint or explanation, not an answer, would be appreciated.
     
  5. Nov 16, 2006 #4
    post your equations, lets see why you get "wierd answers".

    i think anything beyond use mgh, [tex]\frac{mv^2 }{2}[/tex] and [tex]W=U_{initial}-U_{final}[/tex] will be just solving the problem for you...
     
  6. Nov 16, 2006 #5
    Well, for part 1, I thought that I would use
    (MGHf+(1/2)MVf^2)-(MGHi+(1/20MFi^2)
    Then, assuming the block starts at rest, you would get
    ((2kg)(9.8m/s2)(0m)+(1/2)(2kg)(Vf^2))-((2kg)(9.8m/s2)(6m)+(1/2)(2kg)(0m/s))
    But what do I set that equal to? I am just completely lost. My teacher is under review and about to lose his tenure but I still can't bear failing his class. I just need a few pointers. Anything.
     
  7. Nov 16, 2006 #6
    well, [tex]U_{final}-U_{initial}[/tex] equals the energy gained by the system. (and - sign means it lost energy)

    did the system in your question gained\lost any energy?
     
  8. Nov 16, 2006 #7
    Is the work equal to the kinetic energy added to the potential energy?
     
  9. Nov 16, 2006 #8
    oh wait, work is equal to force times displacement, right? so would i just set that whole mess equal to (2kg)(9.8m/s2)(6m) --- (the potential energy)?
     
  10. Nov 16, 2006 #9
    All the information that I have on the problem was given to you in the header. If you are asking me, I would guess that it lost energy to friction?
     
  11. Nov 16, 2006 #10
    no, kinetic energy + potential energy is not work.
    its the total energy of the system (i called it U).
    work is the energy difference between the initial and final states.
     
  12. Nov 16, 2006 #11
    so, is using the U equation ( I think my teacher called it the "energy final-energy initial" formula?) going to find me the initial velocity? I AM SO LOST.
     
  13. Nov 16, 2006 #12
    1) Find the value of the velocity when the block is at the bottom of the plane assuming no friction.

    2) With friction present, the final velocity is only 7.0 m/x at the bottom of the inclined plane. Find work done against friction.

    in the first case you did not have friction - did you lose any energy?

    you solved something much like the second question on a different thread you started...
     
  14. Nov 16, 2006 #13
    Okay, so I think I understand question 2. Question one: I did not lose any energy. So... that would mean that my energy at the beginning of the slide woudl equal that at the end of the slide? So would I set potential energy and kinetic energy equal to each other?
     
  15. Nov 16, 2006 #14
    actually
    I just took the square root of 2*g*h (rad2(9.8)(6)) and got 10.8444m/s. This is because, like you were trying to tell me I think, that it is not changing in energy? It is conservative, right?
     
  16. Nov 16, 2006 #15
    you got it right
     
  17. Nov 16, 2006 #16
    Thank God! And thank you for your help! I won't even ask you for help on the third part, I think I'll just leave it blank, but thanks for your help...
     
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