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    (Original post by Zishi)
    1)The curve starts off with a positive gradient, then approaches zero gradient. How's that not decreasing??

    2)the y-intercept is zero, because then the height fallen = 0, meaning it doesnt have any KE yet
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    (Original post by BrilliantMinds)
    1)The curve starts off with a positive gradient, then approaches zero gradient. How's that not decreasing??

    2)the y-intercept is zero, because then the height fallen = 0, meaning it doesnt have any KE yet
    LOL! I think you mis-read the question. Read part (ii) again, it's asking us to draw the graph of Potential Energy against height - And then read my problem again. :/
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    I think you're misunderstanding them BM - they're being asked to draw another line for GPE.

    You're forgetting air resistance. The ball is actually exchanging its GPE for KE and work against air resistance linearly, so the graph is straight - the reason the KE graph levels off is because a greater proportion of that GPE is work done against air resistance near h0. Also, the curve starts above Ek because the sum of the GPE and KE is higher at first than at h0, because some of that energy has been lost to the air by h0.
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    (Original post by Zishi)
    LOL! I think you mis-read the question. Read part (ii) again, it's asking us to draw the graph of Potential Energy against height - And then read my problem again. :/
    omg, i totally misread it! (Must be the late hour here :P)

    Okay, you are asked to draw GPE against h. Why must be the y-intercept be higher than Ek?

    Total Energy (at start) = GPE

    Total Energy (at end) = KE (max) + heat energy (because while falling through the air the ball excites some air molecules)

    do you see it now?
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    (Original post by Xdaamno)
    I think you're misunderstanding them BM - they're being asked to draw another line for GPE.

    You're forgetting air resistance. The ball is actually exchanging its GPE for KE and work against air resistance linearly, so the graph is straight - the reason the KE graph levels off is because a greater proportion of that GPE is work done against air resistance near h0. Also, the curve starts above Ek because the sum of the GPE and KE is higher at first than at h0, because some of that energy has been lost to the air by h0.
    If air resistance wasn't there, then the graph would've been a straight line, cuz acceleration would've been constant then. And as I explained before, E_p = mgh and the graph is of Energy against Height, so the gradient will be \frac{E_p}{h} and \frac{E_p}{h} = mg. As both g and m would be constant, THEN the graph would be a straight line.
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    (Original post by BrilliantMinds)
    omg, i totally misread it! (Must be the late hour here :P)

    Okay, you are asked to draw GPE against h. Why must be the y-intercept be higher than Ek?

    Total Energy (at start) = GPE

    Total Energy (at end) = KE (max) + heat energy (because while falling through the air the ball excites some air molecules)

    do you see it now?
    Yep. And what about the gradient? Why would it be constant?
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    (Original post by Zishi)
    If air resistance wasn't there, then the graph would've been a straight line, cuz acceleration would've been constant then. And as I explained before, E_p = mgh and the graph is of Energy against Height, so the gradient will be \frac{E_p}{h} and \frac{E_p}{h} = mg. As both g and m would be constant, THEN the graph would be a straight line.
    I'm sorry, I see what you're saying now! But hopefully it's been explained why the initial energy is greater.

    As to the argument that the gradient should not be constant because the gradient = \frac{mgh}{h} and g is affected by air resistance: indeed, the acceleration of the object decreases from g due to the air resistance. However, that's not what the 'g' represents in the context of GPE: the GPE is calculated simply by multiplying the object's weight by its height. For example: an object 2m high falling in viscous fluid with low acceleration has the same GPE as the same object 2m high falling in air.
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    I've modified the graph below to show what would happen with no air resistance.
    In this case, the energy starts as all PE and finishes as all KE.
    The initial PE will be mgh and the final KE will be equal to that amount.
    The two lines will look as shown. The sum of KE plus PE will be constant.
    The lines cross half way down when the PE is 0.5mgh (at half the height) and the KE must be the same.

    With air resistance, the ball does not continue to accelerate at the rate it does at the start.
    It eventually reaches a terminal velocity, shown by the line becoming horizontal.
    The PE line will look the same as before, because PE is always mgh, and its value will be determined by the height the ball is above the base level.
    The difference is that as time goes on, less of this PE gets converter into KE. The difference between the two lines is the amount of energy that has been converted into heat.

    In order to get the slope of the KE line (without air resistance) correct, I've had to extend the graph a bit.
    The PE graph is then determined by this.
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    (Original post by Xdaamno)
    I'm sorry, I see what you're saying now! But hopefully it's been explained why the initial energy is greater.

    As to the argument that the gradient should not be constant because the gradient = \frac{mgh}{h} and g is affected by air resistance: indeed, the acceleration of the object decreases from g due to the air resistance. However, that's not what the 'g' represents in the context of GPE: the GPE is calculated simply by multiplying the object's weight by its height. For example: an object 2m high falling in viscous fluid with low acceleration has the same GPE as the same object 2m high falling in air.
    No problem. So you mean that as we take mg as weight, the gradient remains constant as weight(mg) is constant, right? Thanks anyway! :rolleyes:

    (Original post by Stonebridge)
    I've modified the graph below to show what would happen with no air resistance.
    In this case, the energy starts as all PE and finishes as all KE.
    The initial PE will be mgh and the final KE will be equal to that amount.
    The two lines will look as shown. The sum of KE plus PE will be constant.
    The lines cross half way down when the PE is 0.5mgh (at half the height) and the KE must be the same.

    With air resistance, the ball does not continue to accelerate at the rate it does at the start.
    It eventually reaches a terminal velocity, shown by the line becoming horizontal.
    The PE line will look the same as before, because PE is always mgh, and its value will be determined by the height the ball is above the base level.
    The difference is that as time goes on, less of this PE gets converter into KE. The difference between the two lines is the amount of energy that has been converted into heat.

    In order to get the slope of the KE line (without air resistance) correct, I've had to extend the graph a bit.
    The PE graph is then determined by this.
    Thank you! I got it.
 
 
 
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