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Exercise about mechanical energy

  1. May 2, 2014 #1
    Hi everyone , this exercise was given in one of my midterms , but we didn't correct it and I'm wondering where I went wrong on it : Help will be extremly appreciated :

    Here is the statement :

    A block of mass m=2 kg is pushed by a spring with a spring constant of k=650 N/m which is intially compressed by Δx=0.12m and attached to a wall . This mass slides a distance d= 0.5 m up a frictionless tables wich makes an angle θ=6° with the horizontal .

    1- What is the expression of the total energy of the mass right when it is pushed by the spring ?

    I wrote : E=K+U = 1/2 m*v(t)² + m*g*d(t)*sin θ + 1/2*k*Δx²
    The professor wrote on my paper that 1/2 m*v² =0 (WHY ?)

    2- What is the expression of kinetic enerfy of the mass when it reaches the edge of the table ?
    I wrote ΔK= Wnet =m*g*h+ 1/2*k*Δx² so kowing that Ki=0
    Kf= m*g*d*sinθ + 1/2*k*Δx²
    The professor said that : Kf= -m*g*d*sinθ + 1/2*k*Δx² (WHY ?)
    3- The mass falls a height h=0.8 m down to the ground : With what speed will the block land on the floor :

    I used question 2 , having vf= sqrt(2*g*h+k/m*Δx² ) but as 2 was apparently not correct I lost points on it too !
    Can anyone help ? THANKS A LOT
     
  2. jcsd
  3. May 2, 2014 #2
    1. What values did you use for v(t), d(t), and Δx?

    2. Is the edge of the table = d = 0.5m ?

    3. When the mass leaves the table, does it have vertical or horizontal velocity?
     
  4. May 2, 2014 #3
    Well , for 1 and 2 : As you see , I did not use any value ,as it asks only for expressions !! and d is the distance gone through !
    3- It goes vertically ! it is on a table suspended 5° to the top !
     
  5. May 2, 2014 #4
    Still not clear what you assumed and where the end of the table is but...

    1. If the spring is compressed and then let go, the initial velocity is zero. Hence, the initial kinetic energy is also zero.

    2. After the mass has traveled up the ramp it has gained potential energy and lost kinetic energy. Hence, the -ve sign for the potential energy term.

    3. Actually the mass has both vertical and horizontal components of velocity.
     
  6. May 3, 2014 #5
    I understood your points in 1 and 2 , but 3 ??
     
  7. May 3, 2014 #6

    haruspex

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    When the mass has travelled a distance 0.5m it reaches the edge of the table and falls off. But just before it falls off it is moving up the slope with a speed you can deduce from the answer to (2). It therefore does not fall vertically.
     
  8. May 3, 2014 #7
    At least not initially.
     
  9. May 3, 2014 #8

    haruspex

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    It will never fall vertically.
     
  10. May 3, 2014 #9
    It will fall vertically at the same time that it maintains a horizontal component of velocity.
     
  11. May 3, 2014 #10

    haruspex

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    "falling vertically" means there is no horizontal motion.
     
  12. May 3, 2014 #11
    Relative to what?
     
  13. May 4, 2014 #12

    haruspex

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    In post #3 physicos wrote
    which means straight down, no horizontal component.
    You corrected this in your own post #4, pointing out that it will also have horizontal motion.
    "Falling vertically", in everyday usage, likewise implies no horizontal motion. It is not the same as saying that the vertical component of its motion is downwards.
    In all cases, this is relative to the apparatus.
     
  14. May 4, 2014 #13
    I got your point , vertical and horizontal velocities were important in projectile motion , but in here how are we supposed to get the final velocity ?
     
  15. May 5, 2014 #14
    I think in every day usage you could say something falls vertically and yet not exclude a horizontal component. Do skydivers jumping from a plane fall vertically? If not, how do they fall then?
     
  16. May 5, 2014 #15
    Can you set up the kinematic equations of motion for the mass as it leaves the table?
     
  17. May 6, 2014 #16

    haruspex

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    No, because that would be tautology. Things don't fall horizontally. You would just say it falls.
     
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