Unlock the Mystery of Kinetic Energy with Expert Guidance - Major Help Needed

[Femme06Fatale]

Kinetic Energy -- Major Help Needed

I'm not even looking for answers ... just for somebody who is really willing to help me out here because I definitely do not understand ANY of this stuff that we're doing right now. Any body, please?

 

[futb0l]

What do you need help on?

 

[Femme06Fatale]

Heh .. this is just the first problem of my homework that I entirely DO NOT understand ...

In 1994, Leroy Burrell of the US set what was then a new world record for the me's 100 m run. He ran the 1.00 X 10^2 m distance in 9.85s. Calculate burrelll's kinetic energy, assuming that he ran w/ a constant speed equal to his average speed. Assume his mass was 75.0 kg!

So when I work this all out. .. I got this number such as 3865.08 J ... but that isn't even possible -- I'm so lost :(

 

[DB]

Do you know $$KE=(1/2)mv^2$$?

 

[Femme06Fatale]

well, I guess that maybe it could be right ... it is just .. why do I have to answer it w/ it being 3.87 X 10 ^3? that kind of confused me ...

 

[DB]

Actually I just did the problem and I believe your answer is right. Don't take my word for it though.

 

[Femme06Fatale]

Maybe I just need to learn how to become more confident with this stuff ... but I just can't visualize it very well ... any suggestions as to how I can start to really visualize this stuff more? One final question for the time being ... how do I convert from MJ to J?

 

[DB]

You did this right?

$$Ke=(1/2)*75kg*v^2$$

$$v_{constant}=\frac{100m}{9.85s}=(10.15228426m/s)^2$$

$$Ke=(1/2)*75kg*103.0688757$$

$$Ke=3865.082839 J$$

Seems right to me

 

[Femme06Fatale]

alright now ... say I'm trying to find the kinetic energy and i have an average speed ... ?

 

[DB]

1 MJ= 1 000 000 J

Maybe you mean an accelerated speed, where you can get the average acceleration:

$$a_{average}=\frac{v_2-v_1}{t_s}$$

$$v_2$$ is the final speed
$$v_1$$ is the initial speed.
t is time in seconds

 

[Femme06Fatale]

Thank you :)