calculusxy
  • calculusxy
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
@Michele_Laino
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
  • Michele_Laino
the gravity is perpendicular with respect to the direction of the race

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
|dw:1437835911067:dw|
calculusxy
  • calculusxy
can we somehow formulate the work using force x distance ?
Michele_Laino
  • Michele_Laino
if distance and force are mutually perpendicular, then the work done by that force is null
Michele_Laino
  • 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
  • Michele_Laino
oops...Jack is...
calculusxy
  • calculusxy
i don't understand how the force is null if it is perpendicular.
Michele_Laino
  • 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
  • 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
  • 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
  • calculusxy
so i tried to work with 2N and 5,000m = 10000J for both. how do i show it to be canceling out?
anonymous
  • anonymous
Weight it perpendicular to the direction he ran.
Michele_Laino
  • 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
  • calculusxy
okay
anonymous
  • anonymous
You'd want to do \(2~\text{N}\cdot 5~\text{km} \cdot \cos(90^\circ )= 0~ \text{J}\).
calculusxy
  • calculusxy
we didn't learn trig yet
calculusxy
  • calculusxy
what about for the power?
Michele_Laino
  • 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
  • calculusxy
so for power can we do 10000J/32 min = 312.5 watts
Michele_Laino
  • Michele_Laino
the power also is null, since the work is null
Michele_Laino
  • Michele_Laino
remember that: power= work/ time
calculusxy
  • calculusxy
oh yes
calculusxy
  • calculusxy
thank you so much! i have a last question and then i am done for the day
Michele_Laino
  • Michele_Laino
ok!
calculusxy
  • 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
  • Michele_Laino
here the average speed is: v=10/30=...m/sec
calculusxy
  • calculusxy
1m/sec?
Michele_Laino
  • Michele_Laino
hint: 10/30= 1/3 =...?
Michele_Laino
  • Michele_Laino
\[10:30 \cong 0.334m/\sec \]
Michele_Laino
  • Michele_Laino
now the requasted power is: \[{\text{P = Force }} \times {\text{ speed = }}1000 \times 0.334 = ...watt\]
Michele_Laino
  • Michele_Laino
requested*
calculusxy
  • calculusxy
334 watts?
Michele_Laino
  • Michele_Laino
that's right!
calculusxy
  • calculusxy
so my answer is 334 watts right?
Michele_Laino
  • Michele_Laino
yes! more precisely: P= 333.34 watts
calculusxy
  • calculusxy
THANK YOU TREMENDOUSLY!!!!!!!!!!!!!!!!
Michele_Laino
  • Michele_Laino
:) :)
calculusxy
  • calculusxy
Ciao
Michele_Laino
  • Michele_Laino
Ciao! Hi!
calculusxy
  • calculusxy
how do you say "bye"
calculusxy
  • calculusxy
addio?
Michele_Laino
  • Michele_Laino
yes! I say "Ciao"
calculusxy
  • calculusxy
oh okay. thank you! learned a little bit of italian as well. couldn't get any better !!
Michele_Laino
  • Michele_Laino
ok! :)
calculusxy
  • calculusxy
A presto
Michele_Laino
  • Michele_Laino
see you soon!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.