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

## 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?

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Bystander
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(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?"

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

Bystander
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How much work do you do getting the bucket to the top of the building?

How much work do you do getting the bucket to the top of the building?
Would that be the 3200?

How much work do you do getting the bucket to the top of the building?
I'm confused with that question actually

Bystander
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Let's start at the beginning. What is the minimum force you must apply to lift the bucket?

SteamKing
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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.

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
Because the final answer is supposed to be 4.08kN*m

Excuse me, KE at top of a Building? isnt that suppose to be Potential Energy??

Excuse me, KE at top of a Building? isnt 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

Bystander
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F=KE at the tope of the building
Find the PE=mgh=(20)(9.8)(20)=784 J
And find the KE=(1/2)(20kg)(4.0^2)=3200 J
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.

• Timmy1221
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

SteamKing
Staff Emeritus
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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