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Homework Help: Need Urgent Help with Problems related to Energy and Work

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data
    1. An automobile has a mass of 800 kg and is considered that the efficiency is 14%. (This is 14% of the energy in the fuel is delivered to the wheels). Find the quantity of gasoline used to accelerate the car from rest to 60 miles/h. Note: In this problem, one gallon of gasoline has 1,300,000,000 Joules.

    2. A particle that has 501 KG has a speed of 25m/s in position A and a kinetic Energy of 40,000 joules in position B. Find the total work done when the particle moves from position A to B.

    3. (and last) A person has 80kg and sustains himself on one end of a chord that has 12 meters and at the other end is struck on a building. The cable has 60 degrees with the horizontal reference. How much work is he doing against gravity?

    2. Relevant equations
    I really have no idea what would work in these, since I've tried everything and I still can't get it.

    3. The attempt at a solution
    As I said above, I've tried everything and it'd be a waste of time to post it here.

    Thanks in advance!
  2. jcsd
  3. Mar 27, 2008 #2


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    Staff: Mentor

    Welcome to the PF. We can only offer you tutorial help if you show us your work so far. We do not do your homework for you.

    So, each of these problems involves work. How is work related to energy? How is energy related to mass and velocity? How would those equations be used to start solving each of these questions?
  4. Mar 27, 2008 #3
    In problem #1 here's what we did:

    Kinetic Energy = 800KG x (26.81m/s)² = 287,510.44J, but we have no idea how to relate this to the gasoline or to the efficiency (since we haven't seen efficiency in class, and our teacher gives us problems above our level, so that's why we're coming here.)

    In problem #2 we tried:

    We think that we need to use the W=FxD formula, but we don't have a distance and we have no idea how to get it, so we're stuck there.

    In problem #3:

    We're stuck at the "How much work is he doing against gravity", since we have no idea how "against gravity" works, but we know that since W=FxD CosΘ, so that'd be F=80Kg x 12 (cos60) which would be 480 N?
  5. Mar 27, 2008 #4
    For 1, 14% efficiency means that in one gallon of gasoline, only 14% of that energy is being used to make the car move, the rest is converted to heat.

    For 2, the change in kinetic energy equals the work done.

    I don't know what you're asking in 3 since it's poorly worded.
  6. Mar 27, 2008 #5
    For number 3, remember the definition of Work. Force times a change in distance. That should be enough, if I'm reading the problem right.
  7. Mar 27, 2008 #6
    But how do I apply the efficiency to the answer in #1, like what is the next step to solve the problem?

    And in #2, I don't really understand what you're telling us to do, I'd really love it if you could please be more descriptive, or give us a list of processes to realize it. Much thanks for helping us. :D

    And lemme try to rewrite #3 in a more understandable fashion:

    A person weighs 80kg and sustains himself on one end of a chord that is 12 meters long and at the other end of the chord is a building. The cable is in a 60 degree angle. How much work is he doing against gravity?

    Hope that helps, and thanks for all the help so fast!
  8. Mar 27, 2008 #7
    Say, for example, I have 5 gallons of gasoline that will provide 20 joules of energy. My car is only 10% efficient, so from this 5 gallons of gasoline, I only get out 2 joules of useful mechanical energy.

    For 2:

    [tex]W = Fd = (ma)d = m\frac{(v_f^2-v_i^2)}{2d}d = \frac{m}{2}(v_f^2-v_i^2)=\Delta EK[/tex]
  9. Mar 27, 2008 #8
    So for #1 the next step should be 287,510.44J x .14 then divide the answer by 1.3 billion?
    For #2 I don't have the distance, so how would I fill that spot, or how do I acquire the distance?
  10. Mar 27, 2008 #9
    Why would you need the distance when you know that the work is the change in kinetic energy?
  11. Mar 27, 2008 #10
    Oh, now I understand. So, that'd mean that the answer should be: 156,562.5J (Kinetic Energy in the first position) - 40,000J (Kinetic energy in 2nd position) = 116,562.5 (what unit should I use? Joules?)
  12. Mar 27, 2008 #11
    Work is always measured in joules, as is energy.
  13. Mar 27, 2008 #12
    Ok, thanks. So that'd leave only questions 1 and 3. :)
  14. Mar 28, 2008 #13
    My advice remains, work is force times a change in distance.
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