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Energy problems

  1. Jan 11, 2004 #1
    Question 1.
    A hammer of mass 500 kg is held 10 m above and is going to collide with a pile of mass 1000 kg. The hammer is dropped with gravity and then after the collision, the hammer and the pile have the common velocity and the pile is driven into the ground 0.2 m below. For the figure, please visit here.

    (a) What is the momentum of the hammer before the collision?
    (b) Find the total kinetic energy before and after the collision?
    Account for the difference.
    (c) Find the energy lost from the moment after collision until the moment that the pile is driven 0.2 m into the ground.
    (d) Hence, find the resistive force.

    My answers:
    (a) First should find the velocity of the hammer before the collision.
    Two ways to find,
    (i) Apply the motion formula, first use s = ut + 1/2 at2 where
    u = 0, a = 10, s = 10 to find t. And then use v = u + at with the t just found to find v which is the velocity of the hammer before the collision.
    (ii) Apply potential energy = kinetic energy gained
    potential energy = mgh = kinetic energy = 1/2 mv2 where
    h = 10, m = 500, g = 10. Then also can find the answer.

    After finding the v, find the momentum of the hammer.

    (b) The total kinetic energy before the collision is just equal to the potential energy of the hammer before impact. To find the total kinetic energy after the collision, should find the common velocity of the hammer and the pile after the collision by the conservation of momentum.
    Using m1 * u = (m1 + m2) * v.
    The difference is because some energy is lost due to the sound energy of the impact and the internal energy gained by the hammer and the pile.

    (c) The energy lost = mgh where m = 1000 + 500 = 1500, h = 0.2,
    g = 10.

    (d) The resistive force = energy lost / h where h = 0.2

    Question 2.
    A man of mass 50 kg is above the water surface by 6 m. Then he is going to jump into the water. Suppose that the average resistive force of the water is 1500 N, what is the depth that the man can dive into the water surface (Neglect the man's height, assume he is a point mass)?

    My answer:
    First should find the potential energy of the man which is equal 3000 J. After that the depth of the man below the water
    surface = 3000 / 1500 = 2 m.
  2. jcsd
  3. Jan 11, 2004 #2

    Doc Al

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    Staff: Mentor

    Careful. The question asks for the energy lost beginning after the collision. The total energy immediately after the collision is the total KE (which you found in part b) plus the PE (since it will drop 0.2 m). All of this energy is "lost": converted to internal energy.
    Measure the man's PE from the lowest point---include his underwater distance.
  4. Jan 11, 2004 #3
    Re: Re: Energy problems

    Why the total energy lost also includes the total KE after the collision? I think that energy should be the energy for the pile and the hammer used to driven into the ground for 0.2 m and because the ground has some resistive force, then that will cause the pile and the hammer to lose some of its KE and that loss is the energy lost.
    Would Doc Al give me more explanations about that and what's wrong with my reasoning?
    And is my answer for part (d) of question 1 right?

    I don't quite understand. The man's PE from the lowest point is also the point when his KE is the largest. And at that point, the man should be just above the water surface, right? So is my answer right?

    Thank you for your attention!
  5. Jan 12, 2004 #4

    Doc Al

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    Staff: Mentor

    Re: Re: Re: Energy problems

    Immediately after the collision, the pile + hammer has a certain amount of mechanical energy (KE + PE). All of that energy is lost (transformed into internal energy) as the pile is driven into the ground. In your original answer you included the PE, but not the KE; you must include both.

    The total energy of the pile+hammer equals its KE (after the collision, which you calculated in part b) plus its PE.
    Your answer would be correct if you use the correct value for energy lost.
    The man's lowest point is not at the water's surface---right? Since he plunges below the surface, you must include that contribution of PE as well. Call the depth "d"; the energy of the man is then PE = mg(6 +d). Set that energy equal to the work done by the resistive force of the water: mg(6 + d) = 1500d. Solve for d.
  6. Jan 12, 2004 #5
    Thank you

    Doc Al, thank you very much for your support.
  7. Jan 12, 2004 #6

    Doc Al

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    Staff: Mentor

    You are welcome!
  8. Jan 12, 2004 #7
    Chemical energy->Mechanical energy?

    I have one more question to ask. If a man climbs a flight of steps to go to a higher place, he has chemical energy, so when he reaches the highest place after climbing the steps, should all his chemical energy be transformed to potential energy? More precisely, is the chemical energy in his body greater than, equal or less than the potential energy gained after he has climbed up to the highest point? Should there be any energy loss when he is climbing the steps?
  9. Jan 12, 2004 #8

    Doc Al

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    Staff: Mentor

    Re: Chemical energy->Mechanical energy?

    The man's chemical energy is transformed into PE plus internal energy. The human body is inefficient: the amount of chemical energy lost is greater than the increase in PE. (Most of it goes to body heat.)
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