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?

- calculusxy

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- calculusxy

@Michele_Laino

- Michele_Laino

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

- Michele_Laino

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

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## More answers

- Michele_Laino

|dw:1437835911067:dw|

- calculusxy

can we somehow formulate the work using force x distance ?

- Michele_Laino

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

- Michele_Laino

so, the work done by the weight force of the bottle is null, then also the work done by the Jackis null

- Michele_Laino

oops...Jack is...

- calculusxy

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

- Michele_Laino

the weight force is not null
what is null is the component of the weight force along the direction of the race

- anonymous

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.

- anonymous

From an energy perspective, we know the energy of the bottle doesn't change, so the net work would have to be zero.

- calculusxy

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

- anonymous

Weight it perpendicular to the direction he ran.

- Michele_Laino

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

- calculusxy

okay

- anonymous

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

- calculusxy

we didn't learn trig yet

- calculusxy

what about for the power?

- Michele_Laino

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

- calculusxy

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

- Michele_Laino

the power also is null, since the work is null

- Michele_Laino

remember that:
power= work/ time

- calculusxy

oh yes

- calculusxy

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

- Michele_Laino

ok!

- calculusxy

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.

- Michele_Laino

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

- calculusxy

1m/sec?

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

now the requasted power is:
\[{\text{P = Force }} \times {\text{ speed = }}1000 \times 0.334 = ...watt\]

- Michele_Laino

requested*

- calculusxy

334 watts?

- Michele_Laino

that's right!

- calculusxy

so my answer is 334 watts right?

- Michele_Laino

yes!
more precisely: P= 333.34 watts

- calculusxy

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

- Michele_Laino

:) :)

- calculusxy

Ciao

- Michele_Laino

Ciao! Hi!

- calculusxy

how do you say "bye"

- calculusxy

addio?

- Michele_Laino

yes! I say "Ciao"

- calculusxy

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

- Michele_Laino

ok! :)

- calculusxy

A presto

- Michele_Laino

see you soon!!

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