A worker lifts a 20.0-kg bucket of concrete from the ground

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Homework Help Overview

The problem involves a worker lifting a 20.0-kg bucket of concrete to the top of a 20.0-m-tall building, with the bucket reaching a speed of 4.0 m/s at the top. Participants are discussing the minimum amount of work done by the worker in this scenario, considering both kinetic and potential energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between work, force, and distance, questioning how to apply kinetic energy in the context of lifting the bucket. There are attempts to calculate work using the equations for kinetic and potential energy, and some participants express confusion about the correct approach.

Discussion Status

There are multiple interpretations of how to calculate the work done, with some participants suggesting to add kinetic energy to potential energy. Others are questioning the arithmetic involved in these calculations and expressing uncertainty about the definitions and relationships between the different forms of energy.

Contextual Notes

Participants are grappling with the assumptions regarding the velocity of the bucket and how it relates to the work done in lifting it. There is also a noted confusion about the calculations leading to the expected final answer.

Timmy1221
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Homework Statement


  1. A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 20.0-m-tall building. The bucket is initially at rest, but is traveling at 4.0 m/s when it reaches the top of the building. What is the minimum amount of work that the worker did in lifting the bucket?

Homework Equations


F= ma
W= Fd
F=KE at the tope of the building

The Attempt at a Solution


Since bucket has KE at the top of the building, would I be finding the KE using KE=(1/2)mv^2 equation and then plugging that final number into my W=Fd equation where d would be 20.0 m?

Help please!
 
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Timmy1221 said:
(snip) ... and then plugging that final number into my W=Fd equation where d would be 20.0 m?

How were you "plugging it in?"
 
Bystander said:
How were you "plugging it in?"
I solved for F in my equation of W=Fd by using the kinetic energy formula. I did (1/2)*(20)*(4^2) and got 160J. I plugged that into the W=Fd as my F. So then I solved doing F=160J and d=20m but I got 3200 N*m instead and that's not the right answer. The answer is 4.08kN*m apparently
 
How much work do you do getting the bucket to the top of the building?
 
Bystander said:
How much work do you do getting the bucket to the top of the building?
Would that be the 3200?
 
Bystander said:
How much work do you do getting the bucket to the top of the building?
I'm confused with that question actually
 
Let's start at the beginning. What is the minimum force you must apply to lift the bucket?
 
You are overlooking the change in potential energy of the bucket between the ground and the top of the building.

You also don't know how the bucket acquired the velocity of 4 m/s (whether it was constant the whole time the bucket was rising or if it was in the last second before the bucket reached the top of the building. The best you can do is calculate the KE of the bucket and add it to the change in potential energy.
 
SteamKing said:
You are overlooking the change in potential energy of the bucket between the ground and the top of the building.

You also don't know how the bucket acquired the velocity of 4 m/s (whether it was constant the whole time the bucket was rising or if it was in the last second before the bucket reached the top of the building. The best you can do is calculate the KE of the bucket and add it to the change in potential energy.
So is this what you're saying I should do?

Find the PE=mgh=(20)(9.8)(20)=784 J
And find the KE=(1/2)(20kg)(4.0^2)=3200 J
And then add PE+KE so that my final answer is 784+3200=3894?
Because the final answer is supposed to be 4.08kN*m
 
  • #10
Excuse me, KE at top of a Building? isn't that suppose to be Potential Energy??
 
  • #11
CaptCoonoor said:
Excuse me, KE at top of a Building? isn't that suppose to be Potential Energy??

So what would I do to get my final answer?

Would I solve for my PE with the given numbers from the problem?

I'm feel like I'm getting nowhere
 
  • #12
Timmy1221 said:
F=KE at the tope of the building

Timmy1221 said:
Find the PE=mgh=(20)(9.8)(20)=784 J
And find the KE=(1/2)(20kg)(4.0^2)=3200 J
And then add PE+KE so that my final answer is 784+3200=3894?
Because the final answer is supposed to be 4.08kN*m

You have the method. Now, you need to double check your arithmetic. Once you've done that, you can point at the source of the 4.08 kNm final answer and laugh loudly.
 
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  • #13
Bystander said:
You have the method. Now, you need to double check your arithmetic. Once you've done that, you can point at the source of the 4.08 kNm final answer and laugh loudly.
I can truly be such an idiot sometimes. Thanks I think I need some sleep
 
  • #14
Timmy1221 said:
So is this what you're saying I should do?

Find the PE=mgh=(20)(9.8)(20)=784 J
And find the KE=(1/2)(20kg)(4.0^2)=3200 J
And then add PE+KE so that my final answer is 784+3200=3894?
Because the final answer is supposed to be 4.08kN*m

You might want to check your arithmetic on the calculation of PE again. IDK how you can multiply 400 * 20 and come out with 784.
Ditto the calculation of the KE. IDK how you get 3200 J for this either.

And what do you think a Joule is equal to? You should also check its definition as well.
 

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