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Gain in Kinetic energy

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A 100 newton force lifts a 5 kilogram object 2 metres. When the force is removed, the object continues to move upwards. Calculate: (a) the work done by the force; (b) the gain in gravitational potential energy (using g = 9.5n/kg); (c) the gain in kinetic energy.

    2. Relevant equations
    Ep = mgh
    Ek = 0.5 * m * v^2
    W = F * d
    F * d = 0.5 * m * v^2
    F*d = mgh (lifted upwards)

    3. The attempt at a solution

    a) F * d = 200J (correct)

    b) Ep = 5kg * 9.8N/kg * 2 = 98J (correct)

    c) Need to find the velocity:

    (sqrt(2*(f*d)/m)) = 4(sqrt5)

    v = 4(sqrt5)


    0.5 * 5 * (4(sqrt5))^2 = 200J

    which is wrong, the correct answer is 102 J

    Where did i go wrong?
  2. jcsd
  3. Jul 2, 2011 #2
    This part I don't understand why you're doing so, can you explain it?
  4. Jul 2, 2011 #3
    Hi Sora, we're not supposed to giving direct answer, but instead guide the student.
    It is written in http://https://www.physicsforums.com/showthread.php?t=414380" [Broken]. :-)
    Last edited by a moderator: May 5, 2017
  5. Jul 2, 2011 #4

    Philip Wood

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    Where you went wrong was to assume that the accelerating force, i.e. the force giving the body KE, was the full 100 N applied. The force which accelerates the body is the net or resultant force. This is the resultant of the 100N force and which other force (acting downwards?)

    If you use the resultant force in your calculation to find v2 you'll get the right answer. But in fact, you're going in a circle. There's no need to find v2. Just use the principle of conservation of energy...
  6. Jul 2, 2011 #5
    sorry about that :/
  7. Jul 2, 2011 #6
    I'm finding the velocity to try to find Ek. I rearranged the forumla to get v (think it's correct).

    Also Sora, could i ask why you took 200-98, it asks for the gain in kinetic energy, wouldn't it be just 200J?

    I understand that work done = increase in Ek + increase in Ep and you can figure out Ek by work - Ep but i would of thought that i could just figure it out by the normal equation (0.5 * m * v2)...

    If anyone could explain this it'd be good, thanks for the responses.
  8. Jul 2, 2011 #7
    You did rearrange the formula correctly, but you used F*d for kinetic energy, which is incorrect - F*d is used to find the total change in energy (work), and you used this correctly to answer part a. In order to use that equation to find velocity, you would have to know its kinetic energy beforehand. So the best way to do this problem is to use the law of conservation of energy.

    200J is the total change in energy, and since the increase in gravitational potential is a nonzero number, the increase in kinetic energy would have to be less that 200J.
  9. Jul 2, 2011 #8
    I used F*d just to find the velocity, which is needed to find the kinetic energy right?

    F * d = 0.5 * m * v^2:

    work done = change in kinetic energy, thats the forumla i used to find v.

    In my mind i thought if i could find V, i could use 0.5 * m * v^2 to find the kinetic energy. But then again the question is asking the gain in kinetic energy which i'm still confused about...
  10. Jul 2, 2011 #9

    Philip Wood

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    Have you found the resultant force yet?
  11. Jul 2, 2011 #10
    No, i didn't think i had to as they gave two similar examples, none of which had found the resultant force.
  12. Jul 2, 2011 #11

    Philip Wood

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    Did you read my previous post? You'll find your method works fine if you use the resultant force instead of 100N. Your method is unnecessarily complicated, but at least it's your own method, and it needs only this correction to give the right answer.

    But when you've done this, you need to see the question as a straightforward example of energy conservation. You feed in 200J of energy. The 5 kg object gains 98 J of gravitational potential energy. What happens to the rest of the 200J ?
  13. Jul 2, 2011 #12
    I understand now the energy conservation, but i don't really see how i could find the resultant force of this question.

    I'll get my head around the energy conservation, thanks for all the responses.
  14. Jul 2, 2011 #13

    Philip Wood

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    Resultant upward force = upward force - downward force
    = 100 N - 5x9.8 N
    = 51 N
    It is this resultant force which does the accelerating, and imparts kinetic energy to the body.
  15. Jul 2, 2011 #14
    Thanks, do you have any links on resultant forces (as I've barely been through them (just mentioned slightly in this book))?

    Thanks again everyone.
  16. Jul 2, 2011 #15

    Philip Wood

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    I don't think there's a lot to be said on resultant forces. Suppose a body is acted on by a 3N force to the West, a 5N force to the East, a 6N force downwards and a 6N force upwards. The resultant will be 2N to the East. When you find the body's acceleration using F = ma, the F you use is this resultant, rather than any of the individual forces.

    The resultant is the vector sum of the individual forces. In the example I gave, it was trivially easy to work it out this vector sum, but in general you have to use a vector polygon or equivalent. For example, forces of 10N to the North and 10N to the West would have a resultant of 14N (that is sqrt (10^2 + 10^2)) to the North West.
  17. Jul 2, 2011 #16
    Thanks again,

    if it's not too much to ask could you explain the two examples you put out?

    i understand why you done 9.8 * 5 (9.8n/kg the gravatation field strength * 5kg (mass) (f=ma?)) correct?

    but why the sqrt (10^2 + 10^2)? I'll probably learn about this later on, but just out of curiosity.

  18. Jul 2, 2011 #17

    Philip Wood

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    Ah. You haven't really met vectors yet? Here's an informal introduction. You walk 10 km North and then 10 km West. How far are you from your starting point, and in which direction from your starting point? By drawing a diagram and applying Pyth's theorem, you'll see that you are (sqrt (10^2 + 10^2)) km North West (that is roughly 14 km North West) from your starting point.

    Clearly the directions of the original 10 km excursions are as important in this calculation as their 'magnitudes' (10 km). We call the whole thing, magnitude together with direction, a vector quantity. 'Directed distances', like 10 km North, are called displacements. The 14 km North West is the resultant of 10 km North and 10 km West.

    But there are many other important vector quantities. Force is one of them. Forces add together just as displacements add together. You can represent the forces of 10 N North and 10 N West by arrows 10 cm long pointing North and West respectively. Place the head of one to the tail of the other. Then an arrow going from the tail of the first to the head of the second will represent the resultant force: 14 N North West.
  19. Jul 2, 2011 #18
    Completely get what you mean now, thanks for everything :P
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