Finding the Weight of a Rope and Bucket

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SUMMARY

The discussion focuses on calculating the work done in raising a bucket of water attached to a rope over a well. The rope is 5 meters long with a uniform density of 1 kg/m, and the bucket weighs 20 kilograms. The relevant equation for work is established as W = ∫ F(x) x dx, where the force F(y) is defined as (Mrope + Mbucket) g. The solution involves integrating the force over the distance the bucket is raised, taking into account the variable length of the rope as it is lifted.

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  • Knowledge of physics concepts such as force, work, and potential energy.
  • Familiarity with uniform density and mass calculations.
  • Basic understanding of gravitational force (g = 10 N/kg).
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This discussion is beneficial for students studying physics and calculus, particularly those working on problems involving forces, work, and integration in real-world applications.

glennpagano
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Homework Statement



A rope is 5 meters long and hangs over the edge of a well. It has a bucket of water weighted 20 kilograms attached to one end. For Simplicity, use g=10N/kg

Suppose the rope has a uniform density of [tex]\rho[/tex] = 1kg/m

Homework Equations



W=[tex]\int[/tex] F(x)xdx


The Attempt at a Solution



I could not figure out what the force would be I would think it would be (Mrope+Mbucket)g
Mass of the rope would be ([tex]\rho[/tex])(v)=Mrope
 
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Welcome to PF!

Welcome to PF! :smile:

(have a rho: ρ and an integral: ∫ :wink:)

Does the question ask for the work done in raising the bucket 5 m ?

If so, remember that the rope doesn't rise 5 m … it stops as it gets to the top.

Use work done = change in PE. :wink:
 
I know I can do change in PE but this is for my calculus and my teacher wants me to use calculus I did some more work and found this so far. F(y)=(Mrope+Mbucket) Mrope=[tex]\rho[/tex]v. Now I have to find change in y for that I got (5-y) Now my equation for work is the integral from 0 to 5 of ([tex]\rho[/tex]y+20)(5-y)
 

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