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Assess the statements: about potential and kinetic energy

  1. May 1, 2016 #1
    1. The problem statement, all variables and given/known data

    (direct translations from Finnish to English to the best of my ability)

    Ascertain whether the statements are correct or false. Give reasoning behind all answers.Correct any false statements. (direct translations from Finnish to English)

    a)the change of an object's potential energy doubles during a lift, when the lifting height doubles.

    b)When the speed of a falling object is half of its final speed, the potential energy of that object has been reduced to half of the original.

    2. Relevant equations
    m*g*h= Epot. potential energy

    1/2 * m * v2= Ekin. kinetic energy

    d= 1/2 * a * t2

    delta t = (delta v) / a



    3. The attempt at a solution

    I just wanted to doublecheck my results, because the physicsbook does not provide answers.

    a) To my eyes it looks like the statement is true. Unless there's something very fishy about the wording of the question.

    I dabbled with the equation mgh=E
    I assumed m=1kg , g=9,81m/s2, h=1m

    Obviously it seems that we must keep the mass and acceleration as constants. We affix the mass, and the acceleration to constant values, and we only change the height value. We compare doubled-height-scenario, with the original scenario. It's probably not the most rigorous proof that's out there, but you could make a chart, of the values, and compute the end result. Then you compare the potential energies to each other.

    assuming that: g=9,81, m=1kg, h=1m = 9,81 Joules = scenario 1

    9,81m/s2 * 1kg * (2m)= 19,62 Joules= scenario 2


    we can see that the increased amount of energy is +100%. This means the energy becomes doubled, as the height doubles, whereas the mass and acceleration due to gravity remain the same. The amount of increase, always is the same +100%, when the height doubles. So, whether the original statement was true or false seems debatable.

    b)
    to my eyes it looks like the statement is false.


    However it took some time for me to use those equations to give a result. (especially this one : d= 0,5 * a * t2)
    I suppose one could say that when the object is held stationary at height of 2m. Then the object has zero kinetic energy and 100% of its supposed potential energy.

    When falling object has reached its maximum velocity just before impact at ground. Then that objects kinetic energy is 100% of its supposed kinetic energy. However, it seems that the potential energy is at minimum of zero. When the object is lying on the floor, The height difference between the object and the floor is pretty much zero. Using the zero-product property from mathematics then the potential energy ought to be zero at least regarding the comparison of elevation between the object and the floor, when the object is directly upon the floor.

    The question especially noted that the falling object's velocity should be half of the "endgame velocity". Original velocity at the beginning is assumed as zero by myself.

    I guess we could calculate kinetic energy such that velocity is (end.velocity)/2
    we assume h=2m, m=1kg, a=9,81m/s2 Epot= 19,62 joules

    we assume that energy is transformed from potential energy into kinetic energy at 100% effectiveness.
    we could then calculate the "endgame-velocity" of falling object
    19,62= 0,5 * 1kg * v2

    v2= 19,62/0,5
    v=6,2641839053 m/s

    But we were tasked with 0,5 * endgame-velocity = (3,1320919527 m/s)

    At this point we could I suppose, find the kinetic energy, with that halved speed, and the same mass (1kg)
    0,5 * 1kg * 9,81m2/s2
    =Ekin. = 4,9050 Joules

    If we compare 4,9050 joules with 19,62 joules. We can see that (4,9050)/ (19,62) = 1/4 = 25%
    This tells us that the kinetic energy has increased from 0% to 25% of the original kinetic energy. I suppose, that the rest of the energy must be 75%. 75%+25% = 100%. Potential energy is 75% of the original energy

    (again assuming 100% energy transformation efficiency from potential to kinetic and between)
    0,75 * 19,62 = 14,715 joules

    I suppose one could double-check that result thusly. It took some effort to find especially one of the equations . ( this one; d= 0,5 * a * t2)

    d t= d v / a
    this gives
    t-0 = (3,1320m/s -0) / 9,81m/s2

    t= 0,3192 seconds

    d= 0,5 * 9,81m/s2 * (0,3192 s) 2

    ca. 0,5 metres distance travelled. Object travels this amount downwards from the original drop height.

    2-0,5m = 1,5m

    Epot= 1kg * 9,81m/s2 * 1,5m= 14,715 joules

    14,715 joules / 19,62 joules = 0,75 = 75%
     
  2. jcsd
  3. May 1, 2016 #2

    PeroK

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    That's correct. For part b) can you show that the PE is always 75% when the falling object reaches half its final speed? Not just for some specific values.

    To help you a little, here's how I would have done part a):

    If an object is lifted a height ##h## the gain in PE is:

    ##\Delta PE_1 = mgh##

    If it is lifted a height ##2h## then the gain in PE is:

    ##\Delta PE_2 = mg(2h) = 2(mgh) = 2(\Delta PE_1)##
     
  4. May 1, 2016 #3
    Indulge me. I think I gave it a fair try according to the physicsforum rules. Would any of that have earned any points?

    My book examples did have problems where quite simply the energy conservation principle was evoked. Total energy ought to be 100% in such cases where there is no efficiency losses e.g. to drag, friction... In so doing, the sum of the potential energy and kinetic energy must be 100% in this example problem. Energy can not be destroyed or increased like that, only transformed from one form to another form.


    Presumably this is how it would be done. unless I made some terrible mistake.

    assume 0,25= 1/2*m*1/4v2 Divide both sides 0,25

    1,0=1/2*m*1/4*4/1*v2

    1,0= 1/2*m*4/4*v2

    In the other direction
    assume 1,0 = 1/2 m * v2 times both sides 1/4

    1/4 = 1/2m * 1/4v2

    1/4= 1/2m * (v2/[squareroot(4)]2) invoking potency of quotient math rule. At least thats the direct translation from Finnish to english.

    1/4= 1/2m *(v/2) 2

    but I dunno, that's what came to my mind.
     
  5. May 1, 2016 #4

    PeroK

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    I'm not sure I follow your logic at all. The question asked to correct any false statements, so you should have written for b):

    When the object reaches half its final speed, its potential energy has been reduced by 25%.

    To prove that you should consider the overall energy equation: PE + KE = constant.
     
  6. May 1, 2016 #5
    That subject was not thoroughly covered in my coursebook at all. In fact I'm currently reading about it now.
     
  7. May 1, 2016 #6
    Well this is getting little bit embarrassing but here goes anyway. Comments and corrections are welcome of course.

    I still think the B statement was wrong. And it does seem that the potential energy in the new speed situation (velocity = v/2) does diminish by 1/4. Which would be 25%


    I suppose the idea was little bit of intuitive reasoning, especially with regard to the kinetic energy component. Please tell if this is any better, I suppose.

    We know that immobile objects have zero kinetic energy and also zero velocity. I suppose we can assume that the object is suspended in the beginning at some height = h. Presumably this object is immobile at this moment. Then the object is dropped and the max speed at the bottom only reaches v1/2. V1 by itself denotes the max speed, which is reached by the object in such a case when potential energy has completely transformed into kinetic energy.

    if the speed becomes halved. such as v1/ 2

    then we get

    0,5m * (v/2) (v/2)
    0,5m * v2/4
    0,5m * 1/4 * v2

    However the extra factor 1/4 must be added to the left side of the equation in order to keep it true. It is, as though, both sides of equation are multiplied by 1/4
    1/4 * 1,0= 0,5m * 1/4 * v2
    1/4= 0,5m *(v/2)2


    Originally the equation for 100% kinetic energy would have been
    1,0 = 0,5m * v2

    If the max speed is halved, then the kinetic energy is quartered.

    PEbegin + KEbegin = PEend + KEend

    we know from experience that in normal case max speed = v1
    From here we know also that PEbegin = KEend if the full speed of v1 is reached.
    we also know that KEbegin = 0
    however, the new requirement was it was required that max speed = v1/2


    we could say that PEbegin = 100% of the total energy (because 100% + 0% = 100%)

    1,0 + 0 = x+ 1/4

    x= 1- 1/4
    x= 3/4

    3/4 * 100% = 75%

    100%- 75% = 25%
     
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