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anonymous
 one year ago
Four people (AD) 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?
anonymous
 one year ago
Four people (AD) 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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433089983896:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x axis is "Velocity (m/s)" and yaxis is "Power (W)" too messy trying to write! haha :P

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1is the motion of the ball at constant speed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i believe so :/ it does not indicate but it says same heavy ball along the track so i am assuming so?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! so what do you think for this problem? I am not sure who would have used the least force :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh so that would be person D?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please wait,I'm pondering again...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait so it is not person D?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.0yes:) so who would it be then?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so time t needed to travel that space d is: \[t = \sqrt {\frac{{2d}}{a}} \] a is the acceleration

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we consider two forces, namely F_1 which produces an acceleration a_1, and F_2 which produces an acceleration a_2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.0i am confused so it is still person D?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, the powers of each forces, are given by the change of kinetic energy over time, so we can write: \[{W_1} = \frac{1}{2}\frac{{mv_1^2}}{{{t_1}}},\quad {W_2} = \frac{1}{2}\frac{{mv_2^2}}{{{t_2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.1so, 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
 one year ago
Best ResponseYou've already chosen the best response.0yay! so person D it is! thank you:D
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