## anonymous one year ago 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?

1. anonymous

|dw:1433089983896:dw|

2. anonymous

x axis is "Velocity (m/s)" and y-axis is "Power (W)" too messy trying to write! haha :P

3. Michele_Laino

I'm pondering...

4. anonymous

:)

5. Michele_Laino

is the motion of the ball at constant speed?

6. anonymous

yes i believe so :/ it does not indicate but it says same heavy ball along the track so i am assuming so?

7. Michele_Laino

I think so

8. anonymous

ok! so what do you think for this problem? I am not sure who would have used the least force :/

9. Michele_Laino

10. anonymous

ok!

11. 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

12. anonymous

ohh so that would be person D?

13. Michele_Laino

yes!

14. anonymous

yay!! thank you!!

15. Michele_Laino

please wait,I'm pondering again...

16. anonymous

ooh okie!

17. anonymous

wait so it is not person D?

18. 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?

19. anonymous

yes:) so who would it be then?

20. Michele_Laino

so time t needed to travel that space d is: $t = \sqrt {\frac{{2d}}{a}}$ a is the acceleration

21. 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

22. Michele_Laino

so we can write, since d is the same in both cases: ${t_1} = \sqrt {\frac{{2d}}{{{a_1}}}} ,\quad {t_2} = \sqrt {\frac{{2d}}{{{a_2}}}}$

23. anonymous

ok!

24. anonymous

i am confused so it is still person D?

25. anonymous

@Michele_Laino ?

26. Michele_Laino

now, the powers of each forces, are given by the change of kinetic energy over time, so we can write: ${W_1} = \frac{1}{2}\frac{{mv_1^2}}{{{t_1}}},\quad {W_2} = \frac{1}{2}\frac{{mv_2^2}}{{{t_2}}}$

27. anonymous

ok!

28. Michele_Laino

now we can simplify those formulas as follows: ${W_1} = \frac{{F_1^{3/2}{d^{1/2}}}}{{\sqrt 2 m}},\quad {W_2} = \frac{{F_2^{3/2}{d^{1/2}}}}{{\sqrt 2 m}}$

29. 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!

30. anonymous

yay! so person D it is! thank you:D

31. Michele_Laino

thank you!! :)