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calculusxy
 one year ago
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
 one year ago
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that if the velocity is constant, along each race, then the work done is null, in both races

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the gravity is perpendicular with respect to the direction of the race

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437835911067:dw

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0can we somehow formulate the work using force x distance ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if distance and force are mutually perpendicular, then the work done by that force is null

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, the work done by the weight force of the bottle is null, then also the work done by the Jackis null

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops...Jack is...

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0i don't understand how the force is null if it is perpendicular.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the weight force is not null what is null is the component of the weight force along the direction of the race

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0From an energy perspective, we know the energy of the bottle doesn't change, so the net work would have to be zero.

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0so i tried to work with 2N and 5,000m = 10000J for both. how do i show it to be canceling out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Weight it perpendicular to the direction he ran.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please, note that the writing 2*5000= 10,000 has no meaning, since the weight force and the direction of the race are mutually nperpendicular

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You'd want to do \(2~\text{N}\cdot 5~\text{km} \cdot \cos(90^\circ )= 0~ \text{J}\).

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0we didn't learn trig yet

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0what about for the power?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it 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
 one year ago
Best ResponseYou've already chosen the best response.0so for power can we do 10000J/32 min = 312.5 watts

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the power also is null, since the work is null

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1remember that: power= work/ time

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much! i have a last question and then i am done for the day

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0a new conveyor system at the local packaging plant will use a motorpowered 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
 one year ago
Best ResponseYou've already chosen the best response.1here the average speed is: v=10/30=...m/sec

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: 10/30= 1/3 =...?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[10:30 \cong 0.334m/\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now the requasted power is: \[{\text{P = Force }} \times {\text{ speed = }}1000 \times 0.334 = ...watt\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0so my answer is 334 watts right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! more precisely: P= 333.34 watts

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0THANK YOU TREMENDOUSLY!!!!!!!!!!!!!!!!

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0how do you say "bye"

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I say "Ciao"

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0oh okay. thank you! learned a little bit of italian as well. couldn't get any better !!
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