anonymous
  • anonymous
Four people (A-D) push the same heavy ball 100 m along a track. According to the graph (will draw inside!) who used the least force? person A, person B, person C, or person D?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1433089983896:dw|
anonymous
  • anonymous
x axis is "Velocity (m/s)" and y-axis is "Power (W)" too messy trying to write! haha :P
Michele_Laino
  • Michele_Laino
I'm pondering...

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More answers

anonymous
  • anonymous
:)
Michele_Laino
  • Michele_Laino
is the motion of the ball at constant speed?
anonymous
  • anonymous
yes i believe so :/ it does not indicate but it says same heavy ball along the track so i am assuming so?
Michele_Laino
  • Michele_Laino
I think so
anonymous
  • anonymous
ok! so what do you think for this problem? I am not sure who would have used the least force :/
Michele_Laino
  • Michele_Laino
please wait...
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
I think that there are many variables which can determine the motion of the ball, nevertheless, the minimum force is associated to the minimum power
anonymous
  • anonymous
ohh so that would be person D?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yay!! thank you!!
Michele_Laino
  • Michele_Laino
please wait,I'm pondering again...
anonymous
  • anonymous
ooh okie!
anonymous
  • anonymous
wait so it is not person D?
Michele_Laino
  • Michele_Laino
here is my reasoning: space d traveled by the ball is given by the subsequent formula: \[d = \frac{1}{2}a{t^2}\] since the ball perform an accelerate motion, right?
anonymous
  • anonymous
yes:) so who would it be then?
Michele_Laino
  • Michele_Laino
so time t needed to travel that space d is: \[t = \sqrt {\frac{{2d}}{a}} \] a is the acceleration
Michele_Laino
  • Michele_Laino
now we consider two forces, namely F_1 which produces an acceleration a_1, and F_2 which produces an acceleration a_2
Michele_Laino
  • Michele_Laino
so we can write, since d is the same in both cases: \[{t_1} = \sqrt {\frac{{2d}}{{{a_1}}}} ,\quad {t_2} = \sqrt {\frac{{2d}}{{{a_2}}}} \]
anonymous
  • anonymous
ok!
anonymous
  • anonymous
i am confused so it is still person D?
anonymous
  • anonymous
@Michele_Laino ?
Michele_Laino
  • Michele_Laino
now, the powers of each forces, are given by the change of kinetic energy over time, so we can write: \[{W_1} = \frac{1}{2}\frac{{mv_1^2}}{{{t_1}}},\quad {W_2} = \frac{1}{2}\frac{{mv_2^2}}{{{t_2}}}\]
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
now we can simplify those formulas as follows: \[{W_1} = \frac{{F_1^{3/2}{d^{1/2}}}}{{\sqrt 2 m}},\quad {W_2} = \frac{{F_2^{3/2}{d^{1/2}}}}{{\sqrt 2 m}}\]
Michele_Laino
  • Michele_Laino
so, if we consider their ratio, we get: \[\Large \frac{{{W_1}}}{{{W_2}}} = {\left( {\frac{{{F_1}}}{{{F_2}}}} \right)^{3/2}}\] now if F_1 is less than F_2, then also W_1 is less than W_2, so our statement is proved!
anonymous
  • anonymous
yay! so person D it is! thank you:D
Michele_Laino
  • Michele_Laino
thank you!! :)

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