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Homework Help: Dynamics Question

  1. Apr 4, 2005 #1
    hope someone can help me on this question:

    A 3.0L (3.0kg) bottle of water is thrown vertically upward with a force of 55N. How high will it go if the force is applied for a distance of 80cm starting from rest?
    This answer given is 69cm

    I know that the acceleration of the bottle while the upward force is applied is 8.5m/s^2, and the applied force is 55N, and drew an FBD
    Other than that, i don't know where to start to carry out this question!
  2. jcsd
  3. Apr 4, 2005 #2


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    Look at the equations of motion at constant acceleration.

    v(t) = v_0 + at

    x(t) = x_0 + vt +0.5at²
  4. Apr 4, 2005 #3
    sorry, but i still don't understand
    in those formulas, an amount of time is required isn't it?
  5. Apr 4, 2005 #4


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    First of all, I would like to edit an error in my first post: The equations are

    v(t) = v_0 + at

    x(t) = x_0 + v_0t +0.5at²

    Those are the equations that tell you how the speed and the position of a particle evolve in time when it is subject to a constamt acceleration a.

    You know it starts from rest (v_0 = 0). You can also set your coordinate axes such that x_0 = 0 too. Additionally, F=ma ==> a = F/m = 55/3. Now the equations become considerably simpler:

    v(t) = (55/3)t

    x(t) = 0.5(55/3)t²

    From those, can you find at which time t the bottle is at position x = 0.80 m ?
  6. Apr 4, 2005 #5
    EDIT: Forgot to include work done by gravity.

    You have a force acting over a distance. That means work is done on the system.

    Work done, [tex] U = Fd - mgh = (55 N)*(.8m) - (0.8m)(9.81)(3) = 20.5 J [/tex]

    Work and Energy Equation [tex]T_i + U = T_f [/tex]

    T is kinetic energy [tex] T = \frac{1}{2}mv^2 [/tex]

    [tex] T_i = 0 [/tex] because you start from rest,[tex] v = 0[/tex], so [tex]T = 0[/tex]

    [tex]T_f[/tex] will equal the work done. [tex]T_f = 20.5[/tex] J

    From that you can get speed. [tex]T_f = \frac{1}{2}mv^2 = 20.5[/tex]

    You have mass m, solve for velocity v. [tex]v = 3.69[/tex] m/s

    Now it becomes a kinematics problem. Initial velocity is 3.69 m/s.
    Final velocity is 0 (corresponding to its max height it will reach). Acceleration is -9.81 m/s^2 (gravity)

    Using[tex] v_f^2 = v_i^2 + 2ad [/tex]
    [tex]0 = 3.69^2 + 2(-9.81)(d) [/tex]
    Solve for d
    d = 0.69 m or 69 cm
    Last edited: Apr 4, 2005
  7. Apr 4, 2005 #6
    thx for your help! but i have just one more question.
    The acceleration that was used was -9.81mm/s^2, however, i was just wondering about the acceleration, 8.5m/s^2, of the bottle while the upward force is applied. does that have any affect on this problem? can it be used to solve the problem another way? just wondering :bugeye:
  8. Apr 4, 2005 #7
    Yes, sure.
    Applying [tex]v_f^2 = v_i^2 + 2ad[/tex]
    Let's look at the point from x = 0 to x=0.8. This is where the applied force is being exerted.
    Your FBD should have the force of gravity and the applied force. Summing F = ma you get a = 8.5 m/s^2.

    [tex]v_i = 0[/tex]
    [tex]v_f = \sqrt{ 2 (8.5)(0.8)} = 3.688 m/s[/tex]

    The moment you let go of the bottle and let it fly upward, you have a velocity of 3.688 m/s.
    Now the applied force is no longer there when you let go of the bottle (throwers hand applies the force I assume), only the force of gravity acts on the bottle.
    Applying [tex]v_f^2 = v_i^2 + 2as[/tex]
    [tex] 0^2 = 3.688^2 + 2(-9.81)(s)[/tex]
    s = .69m
    Last edited: Apr 4, 2005
  9. Apr 4, 2005 #8
    Sorry if you read my original first solution. It was wrong. I forgot to include the work done by gravity. Anyways, I made the changes. There are two approaches. Both of which have the same answer. The second seems shorter. But often using work and energy can simplify things which is why I attempted that first.
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