Question

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?

Answer 1

@Michele_Laino

Answer 2

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

Answer 3

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

Answer 4

can we somehow formulate the work using force x distance ?

Answer 5

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

Answer 6

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

Answer 7

oops...Jack is...

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

Weight it perpendicular to the direction he ran.

Answer 14

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

Answer 15

okay

Answer 16

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

Answer 17

we didn't learn trig yet

Answer 18

what about for the power?

Answer 19

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

Answer 20

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

Answer 21

the power also is null, since the work is null

Answer 22

remember that:
power= work/ time

Answer 23

oh yes

Answer 24

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

Answer 25

ok!

Answer 26

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.

Answer 27

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

Answer 28

1m/sec?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

requested*

Answer 33

334 watts?

Answer 34

that's right!

Answer 35

so my answer is 334 watts right?

Answer 36

yes!
more precisely: P= 333.34 watts

Answer 37

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

Answer 38

:) :)

Answer 39

Ciao

Answer 40

Ciao! Hi!

Answer 41

how do you say "bye"

Answer 42

addio?

Answer 43

yes! I say "Ciao"

Answer 44

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

Answer 45

ok! :)

Answer 46

A presto

Answer 47

see you soon!!