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Problem with spring + inclined plane + conversation of energy etc

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A ball (considered as a point) of mass m = 1.5kg is launched from a spring with a Δx= 5cm. Find the elastic constant K knowing that the ball is forced to travel the path ABC and that at point C its speed is zero. AB has no friction while BC has a friction μk=0.05.
    At point C the ball falls downwards with an air resistance of B = 0.5 N*s/m. What is the weight measured by a scale at point E? What is the distance EF?

    This is the drawing
    http://imageshack.us/a/img191/1360/55sp.jpg [Broken]

    Relevant equations
    E0 Wnc = Ef
    Ff = -βv


    3. The attempt at a solution
    I'm not sure how to actually solve this problem, I'm not sure about the process.
    I assumed that at A there's an elastic potential energy, at B gravitational potential energy + kinematic energy and at C all the energy has been "consumed" by the work done by the force of friction since the speed is 0.

    So I tried to set something like this

    1/2kx2 + Wf = mgh + 1/2mv2

    Is this correct? At this point I would solve for K but I don't have the speed for v2[/SUP. Plus where is the kinematic energy? Where does it start? Also I don't understand if to find K I must consider the all ABC path or AB is sufficient. Can you also help find out how to solve the other parts with air resistance and that curved path at the end? Even if you don't wanna give the answer at least knowing the equations to use and laws behind them.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 14, 2013 #2
    I haven't worked the problem but here are some things to consider.
    The vertical component of the initial velocity at A will be converted into potential energy at B.
    The horizontal velocity at A will be converted into frictional work as the mass slides from B to C

    However, I wonder if the mass will stay on the incline and then not overshoot the BC section???
     
  4. Jul 15, 2013 #3
    It will but at point C the speed is 0. I think that my professor made it that way to make things easier for the free fall part also for knowing that friction used all the energy.
    So I don't have to consider that kinect energy?
    Initial Energy: 1/2kx2
    Final Energy: mgh (there's no kinetic energy at C)

    So the equation would become

    1/2kx2 -Wnc = mgh

    I put a negative sign since the work done by friction is on the opposite side of the direction.
    Does this seem correct?

    I just noticed I wrote conversation instead on conservation in the thread title :D
     
  5. Jul 15, 2013 #4
    "1/2kx2 -Wbc = mgh" This seems correct to me.
     
  6. Jul 15, 2013 #5
    I get a value of K = 27048N/m...isn't this too high? maybe wrong calculations
     
  7. Jul 15, 2013 #6
    That is what I got also.
     
  8. Jul 15, 2013 #7
    OK, sorry but this is the first time I use elastic potential energy. Any help about the following questions? I have no idea how to proceed. And thank you"!
     
  9. Jul 15, 2013 #8
    If the mass is going horizontally at point E, then the weight should be mg. Now you have to find the velocity at E in order to find out how far the mass will travel to point F. If it were not for the factor, " = 0.5 N-sec/m you could use the change in potential energy from C to E to calculate the velocity. But since you have this drag factor, I would think you would have to do a little integration to find the work done on the air as the mass dropped. Is this something you are familiar with?
     
  10. Jul 15, 2013 #9
    I'm a little familiar with the drag force, so I have to find the velocity at point E? Is weight measured affected at all by the velocity? I find it odd the professor i just asking for a simple "m times g" :D I think the part I can't really solve is the curved path where I'm given the R (i suppose radius?) of the circumference. Will the speed change from D to E?
     
  11. Jul 15, 2013 #10
    Getting complicated, yes? If the weight is measured on the horizontal part, then the velocity will not matter. However, if the weight is on the curved part, then centripetal force would matter. Another question is whether the drag force exists between D and R or just between C and D?

    I expect one of the physics experts to step in here shortly.
     
  12. Jul 15, 2013 #11
    It's only between C e D then the ball will follow the curved path with no external force.
    In what way does centripetal force affect the speed of the object when it goes to E?
     
    Last edited: Jul 15, 2013
  13. Jul 15, 2013 #12
    Then you will have to find the velocity of the mass at D taking into account the drag. Now, the question is, is the weight measured just before or after the mass passes the transition point whee it is on the curved surface or the horizontal surface. The difference will be the centripital force that goes away after the mass starts traveling horizontally.
     
  14. Jul 15, 2013 #13
    Reviewing the drawing it seems like the scale is positioned at the end of the curved surface so the centripetal force will matter to measure the weight. Is that what you meant? Basically I'm getting that by weight the question is asking what is the F of the centripetal force. Weight = Force = ma = mv2/r. Did I get it? ;)

    The question remains, if I know the initial velocity at D how do I find the final velocity at E to get the Weight (force) and finally the final path EF? something to do with the 90° angle?
     
  15. Jul 15, 2013 #14
    I think the weight just before the mass is traveling horizontal is the centripetal force, mv^2/r, plus mg then after going horizontal, the weight is just mg.

    As to the velocity at D, I am still trying to figure this out. It seems like you might have a differentiable equation to solve unless you already have a formula available.
     
  16. Jul 15, 2013 #15
    humm I think I just got it wrong. I thought that having a centripetal acceleration would increase the tangential velocity from D to E but that actually keeps it constant ( uniform circular motion) so If I have the initial velocity at D (considering the drag force) so the weight at E would the centripetal force + mg, like you said!
     
  17. Jul 15, 2013 #16
    So to find the velocity at D, you must find an expression for he velocity as a function of time, then do an integration to find distance as a function of time, set this equal to the distance of 2m, find the time, plug in and get the velocity. There may be other ways but this is what I did.
     
  18. Jul 15, 2013 #17
    Thank you! about the curved path, doesn't sliding a circular path downwards increase its Angelita speed due to gravity?
     
  19. Jul 15, 2013 #18
    Yes it does, but you can use the change in potential energy between point D and E to determine the increase in kinetic energy from D to E and once you have the KE you can find the velocity.
     
  20. Jul 15, 2013 #19
    I'm having problems with finding the velocity at D. At class we've been told what the drag force is and how to find the terminal velocity of a point mass object which g/Cd that would equal to 19.6m/s. But how do I get the speed after 2m knowing the terminal velocity?
     
  21. Jul 15, 2013 #20
    I am not sure how you learned this topic but what we have here is a case where the drag is proportional to the velocity. This is not always the case and often the drag is proportional to v^2.

    In this case, the velocity of a dropped object can be written as v = (mg/B)(1-e^(-tB/m))
    I hope this looks famaliar. This shows the velocity of a dropped object starting a 0 and approaching exponentially a terminal velocity of (mg/B).

    So, you need a relationship between the velocity and distance as the mass drops 2 meters. What I did was to integrate the velocity to get distance, paying attention to the constant of integration. Now I have distance as a function of time and can fine the time to drop 2 meters. Knowing the time, I can plug back into the velocity equation to get the velocity.

    There may be an easier way to get to this point but this will work. If you neglect drag, you can find the velocity at D to be about 6 m/s. With drag it is a little less.
     
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