How much total work is done in raising the ball?

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Homework Help Overview

The discussion revolves around calculating the total work done in raising a ball in a kinetic sculpture, specifically focusing on the forces acting on the ball, including gravitational and frictional forces. The problem involves concepts from mechanics, particularly work and energy.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for work (W=Fd) and the total energy of the sculpture. There are attempts to clarify how to account for both the work done by the conveyor and the work done against friction. Questions arise regarding the energy at a certain height and the implications of opposing forces on the work done.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the work done. Some guidance has been provided regarding the relationship between the work done by the conveyor and the frictional work. However, there is no explicit consensus on the final calculation or the correct approach to the problem.

Contextual Notes

Participants express confusion about the calculations and the definitions of work and energy in this context. There is mention of missing information, such as the need for a calculator, and the complexity of balancing opposing forces in the work done calculation.

Kiko:)
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A 0.075Kg ball in a kinetic sculpture is raised 1.33m above the ground by a motorized vertical conveyor belt. A constant frctional force of 0.350N acts in the direction opposite the conveyor belt. How much total work is done in raising the ball?

This confuses me quite a bit. I understand some of it. Could someone help me please? :)
 
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How would you normally find work done? What is the total energy of the sculpture?
 
rock.freak667 said:
How would you normally find work done? What is the total energy of the sculpture?

Use the formula W=Fd. And that's the problem I'm having is trying to find the total energy of the sculpture.
 
Kiko:) said:
Use the formula W=Fd. And that's the problem I'm having is trying to find the total energy of the sculpture.

the question says

A 0.075Kg ball in a kinetic sculpture is raised 1.33m above the ground

how much energy does it have at 1.33m above the ground?
 
.099?
 
I'm sorry I'm not trying to sound stupid. I'm just trying to understand this question!
 
Kiko:) said:
.099?

I don't have a calculator but it is around that. So this is the total amount of energy needed to lift the object 1.33m above the ground.

To lift the object, the total work done is the sum of the work done in lifting the object and what else?
 
rock.freak667 said:
To lift the object, the total work done is the sum of the work done in lifting the object and what else?

So then .345N force is acting on it and therefore since there is a force acting the opposite way there is no work that is being done on the ball in the kinetice sculpture. But there is work acting on the conveyor belt?
 
Kiko:) said:
So then .345N force is acting on it and therefore since there is a force acting the opposite way there is no work that is being done on the ball in the kinetice sculpture. But there is work acting on the conveyor belt?

right but the 0.345N does frictional work.

So the total work done = work by conveyor + work done by friction.

What is the work done by friction equal to? (see your 2nd post)
 
  • #10
rock.freak667 said:
right but the 0.345N does frictional work.

So the total work done = work by conveyor + work done by friction.

What is the work done by friction equal to? (see your 2nd post)


Use the equation W=Fd or .099?

So what your saying is to find the work by the conveyor and then add that to work done by friction (.350N) which then will equal the total work done in raising the ball?
 
  • #11
Kiko:) said:
Use the equation W=Fd or .099?

So what your saying is to find the work by the conveyor and then add that to work done by friction (.350N) which then will equal the total work done in raising the ball?

You have the right idea, just jumbled.

Total work done= work done by conveyor+ work done by friction.

You know the left side is 0.099. The work done by friction is Fd. So what is 'd' and the frictional work?
 
  • #12
rock.freak667 said:
You have the right idea, just jumbled.

Total work done= work done by conveyor+ work done by friction.

You know the left side is 0.099. The work done by friction is Fd. So what is 'd' and the frictional work?

Ah, so 0.099 = ______ + .350N x 1.33m?

0.099 + .465 = .564J?
 
  • #13
Kiko:) said:
Ah, so 0.099 = ______ + .350N x 1.33m?

0.099 + .465 = .564J?

yes, just change '+' to '-', I forgot to compensate with signs.
 
  • #14
rock.freak667 said:
yes, just change '+' to '-', I forgot to compensate with signs.

so .099 - .465 = 366J? And tha would be how much total work is done in raising the ball.
 
  • #15
Right?
 
  • #16
Kiko:) said:
so .099 - .465 = 366J? And tha would be how much total work is done in raising the ball.

No the previous answer was correct. It was supposed to be this

0.099 = ______ - .350N x 1.33m
 

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