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Work, forces, friction, motion of a block in a vertical circle. (masteringphysics)

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A small block with mass 0.0325 kg slides in a vertical circle of radius 0.475 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.80 N. In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.680 N.

    How much work was done on the block by friction during the motion of the block from point A to point B?

    2. Relevant equations

    K1+U1+W=K2+U2

    W=Fs

    Conserved energy and motion in a circle

    3. The attempt at a solution

    I know that the distance from A to B is 2*r*pi=(19/20)*pi and I can use that as s in W=Fs. I don't know how to find out the friction.
    All help would be very appreciated!
     
  2. jcsd
  3. Oct 5, 2011 #2

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    That's the hard way. Try the easy way: Compare the mechanical energy at A and B.
     
  4. Oct 5, 2011 #3
  5. Oct 5, 2011 #4

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    The equations N ± mg = mv^2/r are correct, but your calculations are not.

    But that's the right idea to find the speed. (Even better is to find the KE, which is what you really want.)

    Again, you'll eventually want to compare the total mechanical energy at A and B to see how much is lost.
     
  6. Oct 5, 2011 #5
    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    Ok I tried finding KE with Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(7.31m/s)^2 - 0.5*0.0325kg*(3.15m/s)^2=0.7071J which is not correct...

    I don't understand what's wrong with my calculations from above because I just plugged in the given variables to the formula.
     
  7. Oct 5, 2011 #6

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    What formula did you use?
     
  8. Oct 5, 2011 #7
    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    You said that my calculations were wrong. I don't understand why because I just put the given variables into the equation N ± mg = mv^2/r.
     
  9. Oct 5, 2011 #8

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    Well, not exactly. Let's see what you actually did:
    The first part is correct: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)

    But this does not equal 0.68 N.

    You essentially used this formula: N = m(v^2/r), when you should have used: N+mg=m(v^2/r).

    You left off the mg.
     
  10. Oct 5, 2011 #9
    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    Okay thank you.

    Now I got: N+mg=m(v^2/r)=0.0325kg(v^2/0.475m)+(9,8m/s^2)*0.0325kg =0.680N+0.3185N=0.9985N so v=3.82 m/s.

    Then I got: Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.31m/s)^2= -0.631J which is still not correct.

    Should I be using a different formula?
     
  11. Oct 5, 2011 #10

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    This is good for point B. Now fix the other equation you used, for point A.
    Two problems:
    - You're still using your incorrect value for the speed at point A.
    - You're neglecting potential energy.
     
  12. Oct 5, 2011 #11
    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    0.0325kg(v^2/0.475m)=3.80-mg=3.4815N so v=7.13m/s.

    Wtot=K2-K1=0.5mv2^2-0.5mv1^2=0.5*0.0325kg*(3.82m/s)^2 - 0.5*0.0325kg*(7.13m/s)^2= -0.589J

    Is this enough for the correct answer? Do I still need to include the potential energy or have I done that already?

    I'm sorry about being slow, I'm just so confused...
     
  13. Oct 5, 2011 #12

    Doc Al

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    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    Good.
    You still need to include gravitational potential energy. Call the potential energy at the bottom position = 0.

    Use this formula, which you had in your first post: K1+U1+W=K2+U2
     
  14. Oct 5, 2011 #13
    Re: Work, forces, friction, motion of a block in a vertical circle. (masteringphysics

    K1+U1+W=K2+U2
    K1+U1+W=K2+mgh
    0.5*0.0325kg*(7.13m/s)^2+0+W=0.5*0,0325*(3.82m/s)^2+0.0325kg*9.8m/s^2*0.95m
    so W= -0.286J

    Thank you ever so much for the help. :)
     
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