Question

Bob has a tree house at a height of 20 ft. When he's feeling adventurous he uses a rope to climb the tree house instead of the ladder. If the weight of the rope is 32 lb, then how much work must Bob do to pull the rope up once he has entered the house?

Answer 1

Work = force x distance

Answer 2

I can't get the right answer. what is it?

Answer 3

We will need to integrate from 0 to 20 ...

Answer 4

I got 865 but that is not one of my answer choices

Answer 5

@claire7 - how did you get 865?

Answer 6

w=fd..what is the right answer?

Answer 7

could you please explain how you got 865?

Answer 8

by converting 32lb to N and multiplying what the conversion was which was 142 anf thenI did w=142(20) I am so confused

Answer 9

you have left out the acceleration due to gravity

Answer 10

A bit messy but this is what you need to do:

Answer 11

F=ma

Answer 12

you also need to convert feet to meters to get the correct units

Answer 13

we need the right formula for the weight, not just use 32,
because that is a function of how much rope is hanging out of the tree house
James' method covers that

Answer 14

To explain my attachment:
You have a small piece of rope, dx long at a vertical distance x from the tree house. How much work is required to move that piece of rope to the tree house?

Well, it's the weight of that piece of rope times the distance covered. How much mass does that piece dx have? Well there 32 lbs in 20 ft, so for a distance dx of rope, it has a mass of
\[ \frac{32}{20} dx \]
Hence the weight of that piece of rope is
\[ \frac{32}{20} g \ dx \]
and therefore the work required to move that piece of rope to the tree house is
\[ \frac{32}{20} g x \ dx \]

Answer 15

Now, add up all of those pieces of work over the length of the rope, and the total work required to get the rope in the tree house is
\[ W = \int_0^{20} \frac{32}{20} gx \ dx \]

Answer 16

@JamesJ - and then convert to the desired units?

Answer 17

Yes, this is in American units. I'm guessing we can and probably should leave it in these units as this was what was given to us. I'm an SI man myself, but I'd roll with these units for now.

Answer 18

we use SI here as standard :-)

Answer 19

Everywhere else in the world my friend. I wish the Americans would get with the program.

Anyway, @claire, make sense?

Answer 20

kind of you are a genius and I am not would the answer be somehwere around 200?

Answer 21

Not quite. Leaving g as g in the integral, what do you get when you integrate?

Answer 22

i.e., what is the anti-derivative of (32/20)gx ?

Answer 23

15.6?

Answer 24

\[ \int_0^{20} \frac{32g}{20} x \ dx = \frac{32g}{20} \int_0^{20} x \ dx \]
What's the value of
\[ \int_0^{20} x \ dx \]

Answer 25

that does not look english to me I have not learned any of that yet

Answer 26

no calculus allowed?

Answer 27

Ok, well you'll get the same result if you think of moving up the center of mass (CM) of the rope. At what distance is the CM from the tree house?

Answer 28

Where is the center of mass, CM, in the rope?

Answer 29

The CM is exactly in the middle of the rope. Given that, how far does the CM have to move vertically to get into the tree house?