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I think that if the velocity is constant, along each race, then the work done is null, in both races

the gravity is perpendicular with respect to the direction of the race

|dw:1437835911067:dw|

can we somehow formulate the work using force x distance ?

if distance and force are mutually perpendicular, then the work done by that force is null

oops...Jack is...

i don't understand how the force is null if it is perpendicular.

so i tried to work with 2N and 5,000m = 10000J for both. how do i show it to be canceling out?

Weight it perpendicular to the direction he ran.

okay

You'd want to do \(2~\text{N}\cdot 5~\text{km} \cdot \cos(90^\circ )= 0~ \text{J}\).

we didn't learn trig yet

what about for the power?

so for power can we do 10000J/32 min = 312.5 watts

the power also is null, since the work is null

remember that:
power= work/ time

oh yes

thank you so much! i have a last question and then i am done for the day

ok!

here the average speed is:
v=10/30=...m/sec

1m/sec?

hint:
10/30= 1/3 =...?

\[10:30 \cong 0.334m/\sec \]

requested*

334 watts?

that's right!

so my answer is 334 watts right?

yes!
more precisely: P= 333.34 watts

THANK YOU TREMENDOUSLY!!!!!!!!!!!!!!!!

:) :)

Ciao

Ciao! Hi!

how do you say "bye"

addio?

yes! I say "Ciao"

oh okay. thank you! learned a little bit of italian as well. couldn't get any better !!

ok! :)

A presto

see you soon!!