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Initial speed of meteorite

  1. Jun 8, 2014 #1
    It's not homework training problems...I have the answers but I get wrong ans.

    1. The problem statement, all variables and given/known data
    problem 1 What was the initial speed of a meteorite of mass 1 kg if a total of 18 MJ of thermal and other forms of energy were produced during its fall through the atmosphere of a planet and its impact on the surface?

    problem 2: A worker pulls up a load of 0.75 t using a pulley through a height of 2 m. He works at a uniform rate of 300 W. How long will he take to pull up the load? (assume no friction, g = 10 m.s-2)


    2. Relevant equations
     s=,5*a*t^2
    F=ma
    v^2=2*a*s

    3. The attempt at a solution
    problem 1
    18MJ =18e^6 N
    F=ma --> 18e^6=1000*a=18000m.s
    Problem 2
    s=5*a*t^2= 2=.5*10*t^2=0,63 s
     
  2. jcsd
  3. Jun 8, 2014 #2

    Simon Bridge

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    This is not correct - the dimensions do not match.

    ... again, the units do not match.

    ... basically, these are conservation of energy problems.
    Try to describe each process in terms of how the energy transforms.

    Problem 1: what kinds of energy does the meteorite start with? Where does it go?
    Problem 2: How much energy is needed to perform the task? At what rate is the energy being supplied?
     
  4. Jun 8, 2014 #3
    thank you for giving me the hint....
    problem 2 solved
    F=ma=0,75e^3*10m.s=7500N
    P=F*v= v=p/F=300/7500N=0,04m.s
    s=v*t= t=s/v= 2/0,04=50 s

    I still need help with the problem 2 thank you
     
  5. Jun 8, 2014 #4

    D H

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    You obtained the correct answer, but not using Simon's suggestion. You don't need to look at force here. You should only be looking at energy and power in the second problem, energy and velocity in the first.

    Also, there's something wrong with the first problem. The numbers don't add up. A 1 kg meteorite will release a minimum of 63 MJ to the atmosphere and the surface.
     
  6. Jun 8, 2014 #5

    Simon Bridge

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    .... which you didn't use? If you do not use the hints, how can I help you?
    Or do you mean about the units not matching up?

    The King is dead - long live the King?
    It's OK - I know what you mean.

    Note: the net force on the load is zero if it moves at a constant velocity.
    What's more important than your equations is your reasoning.
    If you write down what you mean as well as the equation, you are less likely to make mistakes.

    But you made an heroic effort doing it the hard way so:

    energy needed to lift the load is ##mgh## Joules

    energy is being supplied at a rate ##P## Joules per second

    divide J by J/s and you get time (s)

    thus, the time to lift is: ##t=mgh/P##

    ... now plug the numbers in and see if you got the right answer.
    This lesson is most likely about conservation of energy so to get the best out of it you should use conservation of energy even if you know another way.

    Problem 1 you still need a hand?
    Have you tried answering the questions in post #2?
    Do you know what "conservation of energy" is?

    I agree with DH - the problem statement looks incomplete.
     
  7. Jun 8, 2014 #6
    problem 1 solved**
    Ek=0.5*m*v^2
    18MJ=0.5*1*v^2= 18MJ/0,5=v^2
    sqrt(36e^6)=v^2
    v=6000 m.s
    thank you so much so I just get confused which law to use :/
     
  8. Jun 8, 2014 #7
    which you didn't use? If you do not use the hints, how can I help you?
    Or do you mean about the units not matching up?

    I used it after, I found the ans. the hard way.

    The King is dead - long live the King?
    It's OK - I know what you mean.


    Problem 1 you still need a hand?
    I solved it to look at Ek=0.5mv^2

    Have you tried answering the questions in post #2?
    Do you know what "conservation of energy" is?

    yeah I'm actually reading on it

    I agree with DH - the problem statement looks incomplete.[/QUOTE]

    My english is not good, sorry for that.
     
  9. Jun 8, 2014 #8

    D H

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    I would take "initial velocity" to mean velocity at the top of the atmosphere. The meteorite gains energy due to falling through the atmosphere, but loses energy due to atmospheric drag and impact. That 18 MJ includes energy gained due to falling.
     
  10. Jun 8, 2014 #9
    Did I caculated wrong# I mean did I use the wrong law...

    I understand what you saying...can't figure out the law I have to use....but I got the correct answer.. ? explain to me please
     
  11. Jun 8, 2014 #10

    Simon Bridge

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    You have to describe what happens to the energy of the meteorite as it falls. Use words.
    The equations by themselves don't mean all that much.

    You seem to be saying that the energy lost in falling and the impact is equal to the initial kinetic energy.
    As this appears to be the correct answer I have to wonder...

    Consider:
    Without the atmosphere, the meteorite would hit the ground going faster than when it started ...

    This means that the total energy lost in the decent is more than the initial kinetic energy ... which is why I thought the problem was incomplete. Unless I missed something, we also need to know how high the atmosphere goes and how strong the planet's gravity is.

    But it may be that the person setting the problem missed that too ... or I missed something ;)
     
  12. Jun 8, 2014 #11
    thank you :)
    you haven't missed anything..you just opend my eyes to think better.... :) it's a misstake from the examples. teacher has maded
     
  13. Jun 8, 2014 #12

    Simon Bridge

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    That's the idea :)

    It is tricky when the teacher makes a mistake.

    Energy arguments can be a lot clearer that arguments involving forces.
    The trick is to describe the energy transformations in words and then in maths.
    Enjoy.
     
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