Let's say Jack is running in two 5k races (for the uninitiated, 5k races are 5km long). And in the one, Jack carries a water bottle weighing 2N and he finishes in 32 min. IN the second one, he carries the same water bottle, but finishes in 30min. How much work did Jack do by carrying that water bottle in race 1? Race 2? How much power did he exert in race 1? Race 2?

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Let's say Jack is running in two 5k races (for the uninitiated, 5k races are 5km long). And in the one, Jack carries a water bottle weighing 2N and he finishes in 32 min. IN the second one, he carries the same water bottle, but finishes in 30min. How much work did Jack do by carrying that water bottle in race 1? Race 2? How much power did he exert in race 1? Race 2?

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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

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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
so, the work done by the weight force of the bottle is null, then also the work done by the Jackis null
oops...Jack is...
i don't understand how the force is null if it is perpendicular.
the weight force is not null what is null is the component of the weight force along the direction of the race
If he starts at rest, then he does work on the water bottle to speed it up. If he ends at rest, then he does work on the water bottle to slow it down. The work cancels out to zero.
From an energy perspective, we know the energy of the bottle doesn't change, so the net work would have to be zero.
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.
please, note that the writing 2*5000= 10,000 has no meaning, since the weight force and the direction of the race are mutually nperpendicular
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?
it is suffice that you keep in mind that when force and distance are mutually perpendicular, then the work done by that force is zero
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!
a new conveyor system at the local packaging plant will use a motor-powered mechanical arm to exert an average force of 1000N to push large crates a distance of 10 meters in 30 seconds. determine the power output required of such a motor.
here the average speed is: v=10/30=...m/sec
1m/sec?
hint: 10/30= 1/3 =...?
\[10:30 \cong 0.334m/\sec \]
now the requasted power is: \[{\text{P = Force }} \times {\text{ speed = }}1000 \times 0.334 = ...watt\]
requested*
334 watts?
that's right!
so my answer is 334 watts right?
yes! more precisely: P= 333.34 watts
THANK YOU TREMENDOUSLY!!!!!!!!!!!!!!!!
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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
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