Question

verification please. (posted this b4 apologize)

Answer 1

A rope is 40 meters long and (the rope) has the mass of 10kg. It hangs from one end off the tall building (the building is more than 40 meters). How much work is required to pull the rope up to the top of the building?

Solution:
Force = Mass * Acceleration due to gravity
F = 10kg * 9.8m/s^2 = 98 Newtons.

The distance is varying from y=0 to y=40, so our distance is\[\int\limits_{0}^{40}x~dx\]

then
Work = Force * Distance
Work = \[98 \times \int\limits_{0}^{40}x~dx\]

Answer 2

then I get:\[49x^2|^{40}_{0}~=~49 \times (40^2)~=~78,400\]

Answer 3

so my answer is 78,400 joules.

Answer 4

I said the distance is from y=0 to y=40 and then integrated with respect to x. My error... but am I getting the point ?

Answer 5

why are you dividing by 40 ?

Answer 6

we can assume the rope has uniform density

Answer 7

the rope is 10 kg, and has length 40 meters
density is 10 kg / 40 meter = 0.25 kg / meter

Answer 8

linear density* i should say

Answer 9

it also works out using center of mass approach.
the center of mass is at height 20 meters, it has mass 10 kg, and you are moving the
center of mass 20 meters.

work = m* g * h
= 10 * 9.8 * 20 = 1960 Joules

Answer 10

check
$$
\Large \int_{0}^{40}\frac{10}{40}y \cdot 9.8 dy = 1960


$$

Answer 11

I don't know anything about the density, my teacher hasn't mentioned anything about it in class ( -:( )

I was just thinking that W = D * F
And
D= integral of x from 0 to 40,
F= Mass times acceleration = 10 * 9.8

\[W(x)=(9.8 \times 10)\int\limits_{0}^{40}x~dx\]
in class, we didn't talk much about density.
Also, since the acceleration was given in the question, then be there density it would be given to me as well...

Answer 12

the fact that density is not given to me, I guess, means that I don't need to worry about it.

Answer 13

I am not trying to challenge you, I am just looking for understanding.

Answer 14

you said,
the rope is 10 kg, and has length 40 meters
density is 10 kg / 40 meter

But how does this "density" work out with that
F = m * a
W = F * D
?

Answer 15

delta W = 10 kg / (40 meter) * (delta y meter) * 9.8 meter / second ^2 * y meter

Joule has units kg * m^2 /s^2

Answer 16

density * acceleration * distance, basically ?

Answer 17

you can divide the rope into small chunks of width delta y

Answer 18

yes, that is the only part that I get 100%. (Distance is an integral, because it varies)
Kind of struggling on the other parts.

Answer 19

cup ?

Answer 20

well