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10kg Block free falls from 500ft to 2ft into the ground.

  1. Mar 23, 2016 #1
    Hi, may I request help or quick review to ensure that I present an accurate understanding of this concept. I want to move begin learning about the free fall of one block onto another block, but would like to be sure that I understand the physics of work, force, and energy of one block. Thank you.


    1. The problem statement, all variables and given/known data


    A 10 kg block is dropped from a height such that it free falls for 10 seconds before hitting the ground. When the block reaches the ground, it is driven 2 meters into the ground. Let acceleration of gravity be 10m/s^2 and all kinetic energy is transferred into work.

    What is the net force of the block while it is in free fall?

    How much work is done on the block by gravity while in free fall?

    What is the average acceleration of the block as it's being slowed down by the ground?

    What is the average net force of the block while it is being driven into the dirt?

    What is the average force the block applies to the ground?

    What is the average force the ground applies to block?



    2. Relevant equations

    [itex]x_{f}= \frac{1}{2}at^{2}+v_{0}t + x_{0}[/itex]

    [itex] v_f = at + v_{0} [/itex]

    [itex] v^{2}_{f} = v^{2}_{i}+ 2a(x_{f}-x_{0})[/itex]

    F*d = [itex] \frac{1}{2}mv^{2}_{f} - \frac{1}{2}mv^{2}{i} [/itex]


    3. The attempt at a solution

    [itex] v_{f} = (10m/s^2)(10s)+ 0 = 100m/s [/itex]

    [itex] x_{f} = \frac{1}{2}(10)(10)^{2}[/itex] = 500 meters

    F = ma = (10kg)(10m/s^2) = 100N while in free fall

    F * d = F * 500m = [itex] \frac{1}{2}(10kg)(100m/s)^{2} - \frac{1}{2}(10kg)(0m/s)^{2} = 50,000 J [/itex]
    of kinetic energy. Therefore, gravity has done 50,000 Nm (or Joules) of work on the block during free fall for 500 meters.

    Since the block continues to travel into the ground for 2 meters,
    [itex] v_{f}^{2} = v_{i}^{2} + 2a(x_{f}-x_{0}) [/itex]
    [itex] (0m/s)^{2} = (100m/s)^{2} + 2a(2m) [/itex]
    [itex]a = 2500\frac{m}{s^{2}} [/itex]on average

    Average Net Force of the block: F = ma = 10kg(2500m/s^2) = 25,000N

    Since the ground brings the block back to rest, the ground also applies an average force of 25,000N on the block in the upward direction.
     
    Last edited: Mar 23, 2016
  2. jcsd
  3. Mar 23, 2016 #2

    haruspex

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    For the average acceleration while penetrating the ground, your equation assumes constant acceleration, which is not given. However, since you are told the distance of penetration, not the time, you have no choice but to assume it is constant. Knowing the time you could have used ##\Delta v/t##. Many people, including some teachers, seem to think that a valid alternative is ##\Delta E/d##, but unless the acceleration is constant that produces the wrong answer.

    For the last part, did gravity switch off once it hit the ground?
     
  4. Mar 23, 2016 #3
    Thank you for your response.

    If the force of gravity applied to the block is 100N and the Net Force on the block as it is driven into the ground is 25,000N, then the average force of the ground on the block is F_{net} + F_{gravity} = F_{ground} = 25,000N + 100N = 25,100N. Is this the correct description?


    Yes, I happen to be assuming acceleration is constant while penetrating the ground for ease of comprehension. I had thought about how the acceleration would actually change as the block traveled further into the ground, but I'm only able to make wild guesses at this point.

    Also, not sure I completely understand how ΔE/d would produce a different acceleration than Δv/t.

    [vf = at + v0] = 0m/s = (2500m/s^2)t + 100m/s = \frac{100m/s}{2500} = .04 seconds.
    If I were told from the start that the time duration of the block penetrating the ground is t = .04s,
    then [F*t = Δmv ] = F*(.04s) = (10kg)(0m/s) - (10kg)(100m/s) = 10,000Ns
    where F = 10,000Ns/(.04s) = 250,000N
    and F = ma = 250,000N = (10kg)a ,
    arriving at a constant acceleration of 2500m/s^2.

    Now, computing average acceleration by Δv/t = \frac{100m/s - 0m/s}{.04} = 2500 m/s^2 ,

    Therefore, I don't see how being provided time (.04s) vs. distance (2m) would change any of the other factors in this problem.
     
    Last edited: Mar 23, 2016
  5. Mar 23, 2016 #4

    haruspex

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    Yes.
    If the acceleration is constant they produce the same result. To see an example where ΔE/d gives the wrong answer, consider the ground acting as a one-way spring, i.e. it's SHM until max compression, but there's no rebound.
    The true behaviour of soil is somewhere between the two. There's very little resistance initially, but quickly grows to a maximum.
     
  6. Mar 24, 2016 #5
    Thank you. I'm not sure how SHM applies. After looking it up, I know it is called Simple Harmonic Motion, but not yet familiar with the equation or the physics it describes. When it comes to compression of an object as a force is applied to it, I just learned Hook's Law.

    Is it correct to call Hook's Law a linear function?

    Is Hook's Law similar to SHM?

    In the problem of the block free falling into the ground by 2 meters, is it possible to model the force of the ground on the block as a linear function. (even if not entirely realistic?)

    For instance, when the block hits the ground during the first cm of penetration, the force of the ground on the block is minimal. And as the block is drives further into the ground, the force of the ground on the block increases at a constant rate so that by the time the block reaches 2 meters into the ground, the force has reached a maximum.

    All I have thus far is a constant acceleration of 2500m/s^2. This gives me a constant Force of 25,000N.

    My guess would be to first try to graph the linear graph where the y axis is either constant acceleration or constant force and the x axis is time.


    The graph of constant acceleration 2500m/s over .04 seconds resembles a box with area 100, where 100 represents velocity.

    So, if I attempt to make a new linear equation with a non-horizontal slope such that the area under the curve remains 100, then by solving for a the height of a triangle: (1/2)bh=A => (1/2)(.04s)h = 100 => h = 5000 m/s^2 = acceleration

    And since F = ma = (10kg)(5000m/s^2) = 50,000N

    I have a non-constant linear function (with slope m = 1,250,000) that describes the Force of the ground applied to the block for the duration of .04s.

    Would you consider this an accurate representation of one way in which the force of the ground on the block can be modeled. Not necessarily for the physical reality, but more for a basic comprehension of Force varying, in some way, over time, as it penetrates into the ground?



    Also, if I were to do the same thing, but now with a Force vs. Distance graph, I would first have a constant Force of 25,000N over a distance of 2 meters, which forms a box with area 50,000J.

    In generating a new non constant linear function with area 50,000J, I solve for the height of a triangle: (1/2)(2m)h = 50,000J => h = 50,000N

    So the new non-constant linear function that is describing the Force of the ground along the distance of 2m has a slope of m = rise/run = 25,000/1 while the area under the curve remains 50,000J

    Since Hook's Law states [itex]\frac{Applied Force}{Change in Length}[/itex] = k ,

    would it be an example of Hook's Law if the ground is considered to be compressed by the force of the block.

    And if so, would the spring constant (or elastic constant) of the ground be 50,000N/2m = 25,000 = k ?
     
    Last edited: Mar 24, 2016
  7. Mar 24, 2016 #6

    haruspex

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    I'm not suggesting you try to apply SHM in this question. I was just illustrating that if acceleration is not constant then using ΔE/d gives the wrong answer for average force. The question is at fault in not specifying that acceleration should be taken as constant (and thus, you are to find the actual acceleration, not just the average). See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ for more detail.
     
  8. Mar 27, 2016 #7
    Thank you Haruspex
     
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